6th Year Applied Maths Higher Level Kieran Mills Uniform ...

6th YearApplied Maths Higher Level Kieran Mills

Uniform Accelerated Motion

No part of this publication may be copied, reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from The Dublin School of Grinds.

Ref: 6/appmaths/h/km/UAM notes

Looking to maximise your CAO points?Easter is a crucial time for students to vastly improve on the points that they received in their mock exams. To help students take advantage of this valuable time, The Dublin School of Grinds is running intensive, exam-focused Easter Revision Courses. Each course runs for five days (90 minutes per day). All courses take place in Stillorgan, Co. Dublin.

The focus of these courses is to maximise students’ CAO points.

SPECIAL OFFERBUY 2 COURSES GET A 3RD COURSE FREE

To avail of this offer, early booking is required as courses were fully booked last year.

What do students get at these courses?

9 90 minutes of intensive tuition per day for five days, with Ireland’s leading teachers.

9 Comprehensive study notes.

9 A focus on simple shortcuts to raise students’ grades and exploit the critically important marking scheme.

9 Access to a free supervised study room.

EASTERREVISION COURSES

NOTE: These courses are built on the fact that there are certain predicable trends that reappear over and over again in the State Examinations.

FREE DAILY BUS SERVICE For full information on our Easter bus service, see 3 pages ahead.

To book, call us on 01 442 4442 or book online at www.dublinschoolofgrinds.ie

Access to food and beverage facilities is also available to students.

25% SIBLING DISCOUNT AVAILABLE. Please call 01 442 4442 to avail of this discount.

EASTER REVISION COURSE FEES:6TH YEAR & 5TH YEAR COURSES

PRICE TOTAL SAVINGS

1st Course €295 €295 -

2nd Course €180 €475 €115

3rd Course FREE €475 €410

4th Course €70 €545 €635

5th Course €100 €645 €830

6th Course €100 €745 €1,025

7th Course €100 €845 €1,220

8th Course €50 €895 €1,465

9th Course €50 €945 €1,710

3RD YEAR COURSES

PRICE TOTAL SAVINGS

1st Course €195 €195 -

2nd Course €100 €295 €95

3rd Course FREE €295 €290

4th Course €85 €380 €400

5th Course €50 €430 €545

6th Course €50 €480 €690

7th Course €50 €530 €835

8th Course €50 €580 €980

DSOG Easter 2017 8pg A4 FINAL PRINT.indd 2 20/02/2017 13:25

Timetable An extensive range of course options are available over a two-week period to cater for students’ timetable needs. Courses are held over the following weeks:

» Monday 10th – Friday 14th April 2017

» Monday 17th – Friday 21st April 2017

All Easter Revision Courses take place in The Talbot Hotel, Stillorgan (formerly known as The Stillorgan Park Hotel).

BUY 2 COURSES &GET A 3RD COURSE

F R E E !

6th Year Easter Revision CoursesSUBJECT LEVEL DATES TIME

Accounting H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Agricultural Science H Monday 10th - Friday 14th April 10:00am - 11:30am

Applied Maths H Monday 10th - Friday 14th April 8:00am - 9:30am

Art History H Monday 10th - Friday 14th April 2:00pm - 3:30pm

Biology Course A* H Monday 10th - Friday 14th April 8:00am - 9:30am

Biology Course A* H Monday 17th - Friday 21st April 10:00am - 11:30am

Biology Course B* H Monday 10th - Friday 14th April 10:00am - 11:30am

Biology Course B* H Monday 17th - Friday 21st April 8:00am - 9:30am

Business H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Business H Monday 17th - Friday 21st April 8:00am - 9:30am

Chemistry Course A* H Monday 17th - Friday 21st April 8:00am - 9:30am

Chemistry Course B* H Monday 17th - Friday 21st April 10:00am - 11:30am

Classical Studies H Monday 10th - Friday 14th April 2:00pm - 3:30pm

Economics H Monday 10th - Friday 14th April 8:00am - 9:30am

Economics H Monday 17th - Friday 21st April 10:00am - 11:30am

English Paper 1* H Monday 17th - Friday 21st April 8:00am - 9:30am

English Paper 2* H Monday 10th - Friday 14th April 8:00am - 9:30am

English Paper 2* H Monday 17th - Friday 21st April 10:00am - 11:30am

French H Monday 10th - Friday 14th April 10:00am - 11:30am

French H Monday 17th - Friday 21st April 8:00am - 9:30am

Geography H Monday 10th - Friday 14th April 8:00am - 9:30am

Geography H Monday 10th - Friday 14th April 2:00pm - 3:30pm

German H Monday 17th - Friday 21st April 12:00pm - 1:30pm

History (Europe)* H Monday 17th - Friday 21st April 2:00pm - 3:30pm

History (Ireland)* H Monday 17th - Friday 21st April 12:00pm - 1:30pm

Home Economics H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Irish H Monday 10th - Friday 14th April 10:00am - 11:30am

Irish H Monday 17th - Friday 21st April 12:00pm - 1:30pm

Maths Paper 1* H Monday 10th - Friday 14th April 8:00am - 9:30am

Maths Paper 1* H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Maths Paper 1* H Monday 17th - Friday 21st April 8:00am - 9:30am

Maths Paper 2* H Monday 10th - Friday 14th April 10:00am - 11:30am

Maths Paper 2* H Monday 17th - Friday 21st April 10:00am - 11:30am

Maths Paper 2* H Monday 17th - Friday 21st April 12:00pm - 1:30pm

Maths O Monday 10th - Friday 14th April 10:00am - 11:30am

Maths O Monday 17th - Friday 21st April 12:00pm - 1:30pm

Physics H Monday 17th - Friday 21st April 10:00am - 11:30am

Spanish H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Spanish H Monday 17th - Friday 21st April 10:00am - 11:30am

Note: 5th Year students are welcome to attend any of the 6th Year courses above.

* Due to large course content, these subjects have been divided into two courses. For a full list of topics covered in these courses, please see 3 pages ahead.

5th Year Easter Revision CoursesSUBJECT LEVEL DATES TIME

English H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Maths H Monday 10th - Friday 14th April 10:00am - 11:30am

Note: 4th Year students are welcome to attend any of the 5th Year courses listed above.

3rd Year Easter Revision CoursesSUBJECT LEVEL DATES TIME

Business Studies

H Monday 17th - Friday 21st April 2:00pm - 3:30pm

English H Monday 10th - Friday 14th April 10:00am - 11:30am

English H Monday 17th - Friday 21st April 12:00pm - 1:30pm

French H Monday 17th - Friday 21st April 12:00pm - 1:30pm

Geography H Monday 17th - Friday 21st April 8:00am - 9:30am

German H Monday 17th - Friday 21st April 2:00pm - 3:30pm

History H Monday 10th - Friday 14th April 8:00am - 9:30am

Irish H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Maths H Monday 10th - Friday 14th April 8:00am - 9:30am

Maths H Monday 17th - Friday 21st April 10:00am - 11:30am

Maths O Monday 10th - Friday 14th April 2:00pm - 3:30pm

Science H Monday 10th - Friday 14th April 12:00pm - 1:30pm

Science H Monday 17th - Friday 21st April 8:00am - 9:30am

Spanish H Monday 10th - Friday 14th April 2:00pm - 3:30pm

Note: 2nd Year students are welcome to attend any of the 3rd Year courses above.

6th Year Oral Preparation CoursesWith the Oral marking component worth up to 40%, it is of paramount importance that students are fully prepared for these examinations. These courses will show students how to lead the Examiner towards topics they are prepared for. This will equip students with the information they need to maximise their performance in the State Examinations.

FEES: €140

VENUE: The Talbot Hotel, Stillorgan (formerly The Stillorgan Park Hotel)

SUBJECT LEVEL DATES TIME

French H Sunday 12th March 9:00am - 1:00pm

German H Saturday 11th March 9:00am - 1:00pm

Irish H Sunday 19th March 9:00am - 1:00pm

Spanish H Saturday 11th March 2:00pm - 6:00pm

To book, call us on 01 442 4442 or book online at www.dublinschoolofgrinds.ie

DSOG Easter 2017 8pg A4 FINAL PRINT.indd 3 20/02/2017 13:25

Contents: Uniform ACCelerAted motion

seCtion 1 Using the FormUlae ...................................................................2Exercise 1 ..................................................................................6Exercise 2 ..................................................................................8

seCtion 2 sUccessive times and distances ................................................10Exercise 3 ..................................................................................14

seCtion 3 catch-Up problems ....................................................................16Exercise 4 ..................................................................................25

seCtion 4 velocity-time graphs ................................................................28Exercise 5 ..................................................................................39

seCtion 5 Free-Fall ....................................................................................42Exercise 6 ..................................................................................46

leaving cert QUestions 2015-1996 .............................................................49

©The Dublin School of Grinds Page 1 Kieran Mills & Tony Kelly

Sec

tio

n 1

: USi

ng

th

e F

or

mU

la

e

Qu

an

tity

Sy

mb

ol

un

itS

Acc

eler

atio

n a

m s– 2

Initi

al v

eloc

ity

u m

s– 1

Fina

l vel

ocity

v

m s– 1

Tim

e t

sD

ispl

acem

ent

s m

eq

Uat

ion

S oF

mo

tio

n

Ex.

How

long

doe

s it t

ake

the

bus t

o tra

vel f

rom

A to

B?

How

far i

s it f

rom

A to

B?

Sol

Ut

ion

a u v t

= = = =

−

− −

4 10 50

2 1 1

ms ms

ms

?

vuatt

tt=

+=

+=

∴=

5010

440

4 10s

svut

=+

=+

= =

1 2 1 2 1 2

5010

1060

1030

0()

()(

)(

)()

m

vuat

ss

vut

asut

at

=+

=+

=+

......

...(

)[]

().

...(

)[]

...

1 2N

oN

o1 2

1 22

..()[

]

...(

)[]

....(

)[]

3 4 5

No

No

No

vv

uas

tsvt

a tu

22

1 22

2=

+

=−

0 m

v =

50

m s

-1

AB

u =

10

m s

-1

s =

30

0 m

t =

10

s

a =

4 m

s-2

-R

ight

+Le

ft

+a:

acce

lera

tion

–a:

dece

lera

tion


ex

am

ple 1

u =

3 m

s–1, a

= 6

m s–2

, t =

4 s,

s =

?

ex

am

ple 2

u =

6 m

s–1, a

= 1

0 m

s–2, t

= 4

s, v

= ?

ex

am

ple 3

u =

8 m

s–1, v

= 1

5 m

s–1, t

= 6

s, s

= ?

ex

am

ple 4

u =

4 m

s–1, v

= 9

m s–1

, a =

0.5

m s–2

, s =

?

ex

am

ple 5

s = 1

20 m

, v =

10

m s–1

, t =

3 s,

a =

?

ex

am

ple 6

s = 8

m, u

= 6

m s–1

, a =

–2

m s–2

, t =

?


ex

am

ple 1

u =

3 m

s–1, a

= 6

m s–2

, t =

4 s,

s =

?So

lU

tio

n

mat

he

mat

ica

l c

al

cU

lat

ion

S

u =

3 m

s–1

a =

6 m

s–2

t = 4

ss =

?

No

v pr

esen

t. U

se E

quat

ion

(3).

sut

at=

+

=+

=+

=

1 22 1 2

23

46

412

4860(

)()

()(

)

m

ex

am

ple 2

u =

6 m

s–1, a

= 1

0 m

s–2, t

= 4

s, v

= ?

Sol

Ut

ion

mat

he

mat

ica

l c

al

cU

lat

ion

S

u =

6 m

s–1

a =

10 m

s–2

t = 4

sv

= ?

No

s pre

sent

. Use

Equ

atio

n (1

).

ex

am

ple 3

u =

8 m

s–1, v

= 1

5 m

s–1, t

= 6

s, s

= ?

Sol

Ut

ion

mat

he

mat

ica

l c

al

cU

lat

ion

S

u =

8 m

s–1

v =

15 m

s–1

t = 6

ss =

?

No

a pr

esen

t. U

se E

quat

ion

(2).

vuat

=+

=+

=+

=−

610

46

4046

1()

ms

svut

=+

=+

= =

1 2 1 215

86

323

69()

()

() m


ex

am

ple 4

u =

4 m

s–1, v

= 9

m s–1

, a

= 0.

5 m

s–2, s

= ?

Sol

Ut

ion

mat

he

mat

ica

l c

al

cU

lat

ion

S

u =

4 m

s–1

v =

9 m

s–1

a =

0.5

m s–2

s = ?

No

t pre

sent

. Use

Equ

atio

n (4

).

ex

am

ple 5

s = 1

20 m

, v =

10

m s–1

, t =

3 s,

a =

?So

lU

tio

n

mat

he

mat

ica

l c

al

cU

lat

ion

S

s = 1

20 m

v =

10 m

s–2

t = 3

sa

= ?

No

u pr

esen

t. U

se E

quat

ion

(5).

ex

am

ple 6

s = 8

m, u

= 6

m s–1

, a =

–2

m s–2

, t =

?So

lU

tio

n

mat

he

mat

ica

l c

al

cU

lat

ion

S

s = 8

mu

= 6

m s–1

a =

–2 m

s–2

t = ?

No

v pr

esen

t. U

se E

quat

ion

(3).

vu

ass

ss2

2

22

29

42

05

8116 65=

+

=+

=+

=

(.

)

m

svt

ata

a aa a=

−

=−

=−

=−

=−

=−

1 22

1 22

9 2

120

103

312

030

240

609

918

020

()

()

mms−

2

sut

att

ttt

tt

tt

t

=+

=+

−

=−

−+

=−

−=

=

1 22

1 22

2

286

2

86 6

80

42

02(

)(

)

()(

),s

44s


1. u a t s= = = =− −0 4 61 2m s m s s, , , ?

2. u a t v= = = =− −5 3 31 2m s m s s, , , ?

3. u a t s= = = =− −0 4 41 2m s m s s, , , ?

4. u v t s= = = =− −7 15 101 1m s m s s, , , ?

5. u v s a= = = =− −4 10 71 1m s m s m, , , ?

6. v a t u= = = =− −30 6 21 2m s m s s, , , ?

7. u v s a= = = =− −6 3 1 51 1m s m s m, , . , ?

8. s a t u= = = =−35 6 22m, m s s, , ?

9. s u v t= = = =− −40 5 151 1m m s m s, , , ?

10. u v a s= = = =− − −5 10 0 51 1 2m s m s m s, , . , ?

11. u v t a= = = =− −48 12 91 1m s m s s, , , ?

12. s t a v= = = =−30 2 4 2m s m s, , , ?

13. s t v a= = = =−120 3 10 1m s m s, , , ?

14. s a v t= − = = =− −4 2 32 1m m s m s, , , ?

15. s v a t= = = =− −8 7 31 2m m s m s, , , ?

16. s u a t= = = =− −24 5 21 2m m s m s, , , ?

17. s u a t= = = − =− −8 6 21 2m m s m s, , , ?

18. v a t u= = = =− −72 0 5 11 2km h m s, . , min, ?

19. s v a u= = = =− −8 8 31 2m m s m s, , , ?

20. s v a t= − = = =− −6 4 41 2m m s m s, , , ?

ExErcisE 1. Using thE formUlaE


answErs

ExErcisE 11. 72 m

2. 14 m s–1

3. 32 m

4. 110 m

5. 6 m s–2

6. 18 m s–1

7. –9 m s–2

8. 11.5 m s–1

9. 4 s

10. 75 m

11. –4 m s–2

12. 19 m s–1

13. –20 m s–2

14. 4 s

15. 2 83s s,

16. 3 s

17. 2s, 4s

18. –10 m s–1

19. 4 m s–1

20. 3 s


ExErcisE 2. simplE problEms Using thE formUlaE

1. A train starts from rest and accelerates uniformly at 2.5 m s–2 until it reaches a speed of 25 m s–1. Find the distance moved and the time taken for this motion.

2. A car can accelerate from rest to 90 km h–1 in 7.5 seconds. Find its acceleration.

3. In travelling 65 cm along the barrel of a rifle a bullet accelerates from rest to 230 m s–1. Find the accceleration and the time the bullet is in the barrel.

4. A car travelling at 24 m s–1 requires a minimum braking distance of 36 m. What is its deceleration? How long does it take to stop?

5. A car starts from rest with acceleration 4 m s–2. How far does it go in (i) 2 s, (ii) 3 s, (iii) the third second.

6. A body moves in a straight line and increases its velocity from 3 m s–1 to 15 m s–1 uniformly in 6 s. Find the acceleration and the distance travelled.

7. A particle starts with a velocity of 3 m s–1 and accelerates uniformly at 1.5 m s–2. How far does it go in (i) 1 s, (ii) 5 s, (iii) the fourth second.

8. A body is projected from the origin with a velocity of 8 m s–1 and acceleration –2 m s–2. Find (i) the velocity when t = 3 s, (ii) when it comes to instantaneous rest.

9. A particle moves along a straight line between two points P and Q with constant acceleration 0.8 m s–2. Its velocity at Q is 1.2 m s–1 greater than the velocity at P. If the distance PQ is 48 m, find the velocity at P. How long after passing P does it take the velocity to reach 48 m s–1.

10. A car is moving with speed u m s–1. The brakes of the car can produce a constant deceleration of 5 m s–2. It is known that when the driver decides to stop, a period of 2

5 s elapses before the brakes are applied. As the car passes a point O, the driver decides to stop. Find in terms of u the minimum distance of the car from O when the car comes to rest. The driver is approaching traffic lights and is 102 m away when the light changes from green to amber. The lights remain amber for 3 s before changing to red. Show (a) when u < 30 the driver can stop before reaching the lights, (b) when u > 34 the driver can pass the light before it turns red.


answErs

ExErcisE 21. 125 m, 10 s

2. 103

2ms−

3. 40, 692 m s–1, 5 652 10 3. × − s

4. –8 m s–2, 3 s

5. 8 m, 18 m, 10 m

6. 2 m s–2, 54 m

7. 3.75 m, 33.75 m, 8.25 m

8. 2 m s–1, 4 s

9. 31.4 m s–1, 20.75 s

10. 110

2 25u u+


Sec

tio

n 2

: Su

cc

eSS

ive t

ime

S an

d d

iSta

nc

eS

ex

am

ple 1

In tw

o su

cces

sive

seco

nds a

uni

form

ly a

ccel

erat

ing

body

trav

els 5

m a

nd

13 m

. Fin

d its

acc

eler

atio

n an

d its

initi

al v

eloc

ity.

ex

am

ple 2

A u

nifo

rmly

dec

eler

atin

g bo

dy c

over

s suc

cess

ive

100

m d

ista

nces

in 5

s an

d 10

s. F

ind

its in

itial

spee

d, th

e de

cele

ratio

n an

d th

e fu

rther

tim

e fo

r th

e bo

dy to

com

e to

rest

.


ex

am

ple 1

In tw

o su

cces

sive

seco

nds a

uni

form

ly a

ccel

erat

ing

body

trav

els 5

m a

nd 1

3 m

. Fin

d its

acc

eler

atio

n an

d its

initi

al

velo

city

.So

lu

tio

n

a

t =

1 s

t =

2 s

, s =

18

m

t =

1 s

us =

5 m

s =

13

m

sut

at=

+1 2

2

182

218

22

1 22

=+

=+

ua

ua

()

()

....(

)2

51

15 10

2

1 22

1 2

=+

=+

=+

ua

ua

ua

()

()

....(

)1

In th

ese

type

s of p

robl

ems a

lway

s use

the

equa

tion

show

n.

Take

all

valu

es fr

om th

e be

ginn

ing

poin

t.m

ath

em

atic

al c

al

cu

lat

ion

S

An

swer

s

102

118

22

102

182

28

10

=+

×−

=+

−=−

−=

+=

ua

ua

ua

ua

a

......

()(

)...

.()1 2

==+

=+

=∴

=

==

−−

210

28

22

1

81

21

ua

u uu

au....(

)

,

1

ms

ms


sut

at=

+1 2

2

In th

ese

type

s of p

robl

ems a

lway

s use

the

equa

tion

show

n.

Take

all

valu

es fr

om th

e be

ginn

ing

poin

t.

ut =

5 s

t =

15

s, s

= 2

00

m

s =

10

0 m

s =

10

0 m

t =

10

st =

?

a

0 m

s-1

100

55

100

540

25

1 22

25 2

=+

=+

=+

ua

ua

ua

()

()

.....(

)1

200

1515

200

1580

645

1 22

225 2

=+

=+

=+u

au

au

a

()

()

.....(

)2

mat

he

mat

ica

l c

al

cu

lat

ion

S

402

53

806

45

120

615

806

=+

×−

=+

−=−

−=

+

ua

ua

ua

u

......

.()(

)...

..()1 2

44540

30

402

540

25

402

120

62

4 3

20 3

aa

ua

u u u

−=

=+

=+

−

=−

=−

......

.()

()1

0014

06

140 6

70 31

= =

∴=

−

uu

um

s

co

nt...

.

ex

am

ple 2

A u

nifo

rmly

dec

eler

atin

g bo

dy c

over

s suc

cess

ive

100

m d

ista

nces

in 5

s an

d 10

s. F

ind

its in

itial

spee

d, th

e de

cele

ratio

n an

d th

e fu

rther

tim

e fo

r the

bod

y to

com

e to

rest

.So

lu

tio

n


ex

am

ple 2

A u

nifo

rmly

dec

eler

atin

g bo

dy c

over

s suc

cess

ive

100

m d

ista

nces

in 5

s an

d 10

s. F

ind

its in

itial

spee

d, th

e de

cele

ratio

n an

d th

e fu

rther

tim

e fo

r the

bod

y to

com

e to

rest

.So

lu

tio

n

sut

at=

+1 2

2

ut =

5 s

t =

15

s, s

= 2

00

m

s =

10

0 m

s =

10

0 m

t =

10

st =

?

a

0 m

s-1

mat

he

mat

ica

l c

al

cu

lat

ion

S

402

53

806

45

120

615

806

=+

×−

=+

−=−

−=

+

ua

ua

ua

u

......

.()(

)...

..()1 2

44540

30

402

540

25

402

120

62

4 3

20 3

aa

ua

u u u

−=

=+

=+

−

=−

=−

......

.()

()1

0014

06

140 6

70 31

= =

∴=

−

uu

um

s

mat

he

mat

ica

l c

al

cu

lat

ion

S

An

swer

: Fur

ther

tim

e =

17.5

– 1

5 =

2.5

s

Furth

er ti

me

for b

ody

to c

ome

to re

st:

u a v t

= =−

= =

−

−

−

70 31

4 32

10

ms ms

ms

?

vuat

tt t t

=+

=+−

= = ==

0 470

175

70 34 3

4 370 3

70 4

()

.s


ExErcisE 3. succEssivE TimEs/succEssivE DisTancEs

1. In two successive seconds a uniformly accelerating body travels 4 m and 8 m. Find its acceleration.

2. A uniformly accelerating body travels 5 m and 11 m repectively in its first two seconds. How far does it travel in the fourth second?

3. A uniformly decelerating body covers successive 100 m distances in 5 s and 10 s. Find its initial speed, the deceleration and the further time for the body to come to rest.

4. A particle starts from rest and moves in a straight line with uniform acceleration. It passes three points A, B and C where |AB| =105 m and |BC| = 63 m. If it takes 6 s to travel from A to B and 2 s from B to C find (i) its acceleration, (ii) the distance of A from the starting position.

5. A sprinter runs a race with constant acceleration throughout. During the race he passes four posts A, B, C, D such that |AB| = |BC| = |CD| = 36 m. If the sprinter takes 3 s to run from A to B and 2 s to run from B to C, how long does it takes to run from C to D?

6. A particle moving in a straight line with uniform acceleration describes 23 m in the fifth second of its motion and 31 m in the seventh second. Calculate its initial velocity.

7. A body travels in a straight line with uniform acceleration. The particle passes three points A, B and C at t = 0, t = 3 s and t = 6 s. If |BC| = 90 m and the speed of the particle at B is 21 m s–1, find the acceleration of the body and its speed at A.

8. A, B, C are three points which lie in that order on a straight road with |AB| = 45 m and |BC| = 32 m. A car travels along the road in the direction ABC with constant acceleration f. The car passes A with speed u and passes B five seconds later and passes C two seconds after that. Find u and f.

9. A car is moving along at a steady 20 m s–1 when the driver suddenly sees a tree across the road 56 m ahead. He immediately applies the brakes giving the car a constant deceleration of 4 m s–2. How far in front of the tree does the car come to rest? If the driver had not reacted immediately and the brakes were applied one second later with what speed would the car have hit the tree?

10. A, B, C are three points on a straight line in that order. A body is projected from B towards A with a speed of 5 m s–1. The body experiences an acceleration of 2 m s–2 towards C. If |BC| = 24 m, find the time to reach C, and the distance travelled by the body from the instant of projection until it reaches C.

11. A bus 12.5 m long travels with constant acceleration. The front of the bus passes a point P with speed u and the rear passes P with speed v. Find in terms of u and v (i) the time taken for the bus to pass P, (ii) what fraction of the bus passes P in half this time.


12. A body moving in a straight line with constant acceleration passes in succession through points A, B, C and D where |AB| = x, |BC| = y and |CD| = z where the distances x, y and z are covered in equal intervals of time. Show 2y = x + z.

13. A uniformly decelerating train of length 40 m enters a station of length 80 m. The front engine leaves the station 5 s later and the rear of the train leaves the station after a further 5 s. Find the deceleration of the train.

14. A uniformly accelerating body starts with a speed of u, in successive times of t travels

distances s and 2s. Prove that its acceleration is 4 2us

.

15. A body starts moving in a straight line with velocity u and acceleration a. If when the velocity has increased to 5u the acceleration is reversed in direction its magnitude being unaltered prove that when the particle returns to its starting point its velocity will be –7u.

answErs

ExErcisE 31. 4 m s–2

2. 23 m

3. 703

1 43

2 2 5ms ms s− −−, , .

4. (i) 3.5 m s–2, (ii) 7 m

5. 1.6 s

6. 5 m s–1

7. 6 m s–2, 3 m s–1

8. 4 m s–1, 2 m s–2

9. 6 m, 10.6 m s–1

10. 8 s, 36.5 m

11. 25 3

4v uu vu v+++

,( )

13. –1.6 m s–2


Sec

tio

n 3

: cat

ch

-up

pro

bl

em

S

ex

am

ple 1

Two

bodi

es st

art t

oget

her a

t the

sam

e tim

e at

the

sam

e pl

ace

and

mov

e al

ong

the

sam

e st

raig

ht li

ne. I

f one

mov

es w

ith a

con

stan

t spe

ed o

f 16

m s–1

whi

le th

e ot

her s

tarts

from

rest

and

mov

es a

t a c

onst

ant

acce

lera

tion

of 4

m s–2

. How

long

will

it ta

ke b

efor

e th

ey a

re to

geth

er?

ex

am

ple 2

Two

bodi

es A

and

B tr

avel

in th

e sa

me

dire

ctio

n al

ong

the

sam

e lin

e.

Bod

y A

star

ts w

ith v

eloc

ity 5

m s–1

and

acc

eler

atio

n 3

m s–2

. The

oth

er

body

star

ts fr

om th

e sa

me

plac

e w

ith v

eloc

ity 2

m s–1

and

acc

eler

atio

n

4 m

s–2. F

ind

whe

n an

d w

here

they

are

toge

ther

aga

in.

ex

am

ple 3

Two

bodi

es m

ove

in th

e sa

me

dire

ctio

n al

ong

para

llel p

aths

. A st

arts

fr

om a

poi

nt O

with

vel

ocity

8 m

s–1 a

nd a

ccel

erat

ion

2 m

s–2 a

nd B

star

ts

8 m

ahe

ad o

f A a

nd m

oves

off

with

vel

ocity

2 m

s–1 a

t acc

eler

atio

n

4 m

s–2. F

ind

whe

n th

ey w

ill b

e to

geth

er a

nd th

eir d

ista

nces

from

O a

t th

ese

times

. Wha

t are

thei

r res

pect

ive

spee

ds w

hen

they

are

toge

ther

?

A

B

64

m

u =

10 m

s–1

a =

2 m

s–2

u =

1 m

s–1

a =

4 m

s–2

ex

am

ple 4

If A

star

ts 2

seco

nds b

efor

e B

find

whe

n an

d w

here

they

are

toge

ther

. Fi

nd th

e m

axim

um d

ista

nce

that

A m

oves

ahe

ad o

f B in

the

subs

eque

nt

mot

ion.


ex

am

ple 1

Two

bodi

es st

art t

oget

her a

t the

sam

e tim

e at

the

sam

e pl

ace

and

mov

e al

ong

the

sam

e st

raig

ht li

ne. I

f one

mov

es

with

a c

onst

ant s

peed

of 1

6 m

s–1 w

hile

the

othe

r sta

rts fr

om re

st a

nd m

oves

at a

con

stan

t acc

eler

atio

n of

4 m

s–2.

How

long

will

it ta

ke b

efor

e th

ey a

re to

geth

er?

Sol

ut

ion

mat

he

mat

ica

l c

al

cu

lat

ion

S

sut

at=

+1 2

2

An

swer

s

u =

16

m s

a =

0 m

s

-1 -2

s=

sA

B

A B

u =

0 m

sa

= 4

m s

-1 -2

tt

t =

0 s

st

tt

st

tt

A B

=+

= =+

=

160

16 04

2

1 22

1 22

2

()

()

()

ss

tt

tt

tt

tttA

B−

=

−=

−=

−=

−=

=

0

162

02

160

80

80

082

2

2 () ,s

s

sepA

rAt

ion

eq

uAt

ion

: s

sA

B−

tog

eth

er (l

evel

): s

sA

B−

=0

A a

nd B

are

toge

ther

afte

r 0 s

and

afte

r 8 s.

ss

tt

AB

−=

−16

22


ex

am

ple 2

Two

bodi

es A

and

B tr

avel

in th

e sa

me

dire

ctio

n al

ong

the

sam

e lin

e. B

ody

A st

arts

with

vel

ocity

5 m

s–1 a

nd

acce

lera

tion

3 m

s–2. T

he o

ther

bod

y st

arts

from

the

sam

e pl

ace

with

vel

ocity

2 m

s–1 a

nd a

ccel

erat

ion

4 m

s–2. F

ind

whe

n an

d w

here

they

are

toge

ther

aga

in.

Sol

ut

ion

mat

he

mat

ica

l c

al

cu

lat

ion

S

sut

at=

+1 2

2

u =

5 m

sa

= 3

m s

-1 -2

s=

sA

B

A B

u =

2 m

sa

= 4

m s

-1 -2

tt

t =

0 s

st

tt

t

st

tt

t

A B

=+

=+

=+

=+

()

()

()

()

53

5 24

22

1 22

3 22 1 2

2

2

30

60

60

06

1 22

2tt

tt

tt

t

−=

−=

−=

=() ,s

s

sepA

rAt

ion

eq

uAt

ion

: s

sA

B−

tog

eth

er (l

evel

): s

sA

B−

=0

ss

tt

tt

tt

AB

−=

+−

−

=−

52

2

3

3 22

2

1 22

A a

nd B

are

toge

ther

afte

r 0 s

and

afte

r 6 s.

Dis

tanc

e A a

nd B

hav

e tra

velle

d af

ter 6

s:s

tt

B=

+

=+

=+

=+

=

22

26

26

26

236

1272

84

2

2(

)(

)(

)(

)

m


ex

am

ple 3

Two

bodi

es m

ove

in th

e sa

me

dire

ctio

n al

ong

para

llel p

aths

. A st

arts

from

a p

oint

O w

ith v

eloc

ity 8

m s–1

and

ac

cele

ratio

n 2

m s–2

and

B st

arts

8 m

ahe

ad o

f A a

nd m

oves

off

with

vel

ocity

2 m

s–1 a

t acc

eler

atio

n 4

m s–2

. Fin

d w

hen

they

will

be

toge

ther

and

thei

r dis

tanc

es fr

om O

at t

hese

tim

es. W

hat a

re th

eir r

espe

ctiv

e sp

eeds

whe

n th

ey

are

toge

ther

?So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

Ssut

at=

+1 2

2B

star

ts 8

m a

head

of A

. The

refo

re, a

dd

8 to

the

dist

ance

equ

atio

n fo

r B.

u =

8 m

sa

= 2

m s

-1 -2

s=

sA

B

A

B u =

2 m

sa

= 4

m s

-1 -2

o

8 m

tt

t =

0 s

st

ttt

st

tt

t

A B

=+

=+

=+

+

=+

+

()

()

()

()

82

8 82

4

82

2

1 22

2

1 22

2

−+

−=

−+

=−

−=

=tt

tt

tt

t

2

2

68

06

80

24

02

4(

)()

,ss

Co

nt.

....

sepA

rAt

ion

eq

uAt

ion

: s

sA

B−

tog

eth

er (l

evel

): s

sA

B−

=0

ss

tt

tt

tt

AB

−=

+−

−−

=−

+−

88

22

68

22

2


ex

am

ple 3

Two

bodi

es m

ove

in th

e sa

me

dire

ctio

n al

ong

para

llel p

aths

. A st

arts

from

a p

oint

O w

ith v

eloc

ity 8

m s–1

and

ac

cele

ratio

n 2

m s–2

and

B st

arts

8 m

ahe

ad o

f A a

nd m

oves

off

with

vel

ocity

2 m

s–1 a

t acc

eler

atio

n 4

m s–2

. Fin

d w

hen

they

will

be

toge

ther

and

thei

r dis

tanc

es fr

om O

at t

hese

tim

es. W

hat a

re th

eir r

espe

ctiv

e sp

eeds

whe

n th

ey

are

toge

ther

?So

lu

tio

n

sut

at=

+1 2

2vuat

=+

u =

8 m

sa

= 2

m s

-1 -2

s=

s=

20

mA

B

A

B u =

2 m

sa

= 4

m s

-1 -2

o

8 m

v=

A12

m s

-1

v=

B10

m s

-1

t =

0 s

t =

2 s

t =

4 s

mat

he

mat

ica

l c

al

cu

lat

ion

S

Dis

tanc

es a

nd sp

eeds

afte

r 2 s:

stt

vuat

vA A B

=+

=+

=+

= =+

=+

=+

=−

8 82

216

420 8

22

84

12

2

2

1

()

()

()

()(

)

m ms

==+

=+

=+

=−

uat

()

()(

)2

42

28

101

ms

st

ttt

st

tt

t

A B

=+

=+

=+

+

=+

+

()

()

()

()

82

8 82

4

82

2

1 22

2

1 22

2

Co

nt.

....


ex

am

ple 3

Two

bodi

es m

ove

in th

e sa

me

dire

ctio

n al

ong

para

llel p

aths

. A st

arts

from

a p

oint

O w

ith v

eloc

ity 8

m s–1

and

ac

cele

ratio

n 2

m s–2

and

B st

arts

8 m

ahe

ad o

f A a

nd m

oves

off

with

vel

ocity

2 m

s–1 a

t acc

eler

atio

n 4

m s–2

. Fin

d w

hen

they

will

be

toge

ther

and

thei

r dis

tanc

es fr

om O

at t

hese

tim

es. W

hat a

re th

eir r

espe

ctiv

e sp

eeds

whe

n th

ey

are

toge

ther

?So

lu

tio

n

sut

at=

+1 2

2vuat

=+

st

ttt

st

tt

t

A B

=+

=+

=+

+

=+

+

()

()

()

()

82

8 82

4

82

2

1 22

2

1 22

2

mat

he

mat

ica

l c

al

cu

lat

ion

S

Dis

tanc

es a

nd sp

eeds

afte

r 4 s:

stt

vuat

vA A

=+

=+

=+

= =+

=+

=+

=−

8 84

432

1648 8

24

88

16

2

2

1

()

()

()

()(

)

m ms

BBuat

=+

=+

=+

=−

()

()(

)2

44

216

181

ms

u =

8 m

sa

= 2

m s

-1 -2

s=

s=

20

mA

B

A

B u =

2 m

sa

= 4

m s

-1 -2

o

8 m

v=

12

m s

A

-1v

= 1

6 m

sA

-1

v=

10

m s

B

-1v

= 1

8 m

sB

-1

s=

s=

48

mA

B

t =

0 s

t =

2 s

t =

4 s


ex

am

ple 4

If A

star

ts 2

seco

nds b

efor

e B

find

whe

n an

d w

here

they

are

toge

ther

. Fin

d th

e m

axim

um d

ista

nce

that

A m

oves

ahe

ad o

f B in

the

subs

eque

nt m

otio

n.So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

S

sut

at=

+1 2

2

If A

star

ts 2

s be

fore

B, t

ake A

’s ti

me

as (t

+ 2

) s

and

B’s

tim

e as

t s.

u =

10

m s

a =

2 m

s

-1

-2

s=

sA

B

A

Bu =

1 m

sa

= 4

m s-1

-2

64

mt

t +

2

t =

0 s

t =

2 s

−+

−=

−+

=−

−=

=tt

tt

tt

t

2

2

1340

013

400

58

05

8(

)()

,ss

A a

nd B

are

toge

ther

(lev

el) 5

s an

d 8

s afte

r B st

arts

.

sepA

rAt

ion

eq

uAt

ion

: s

sA

B−

ss

tt

tt

tt

AB

−=

++

−−−

=−

+−

22

2

1424

264

1340

tog

eth

er (l

evel

): s

sA

B−

=0

Co

nt.

....

st

tt

tt

tt

t

A=

++

+

=+

++

=+

++

+

=+

()(

)(

)()

()

102

22

1020

210

204

41

1 22

2

2

244

24

641

4

264

1 22

2

t

st

tt

tB

+

=+

+

=+

+

()

()

st

tt

tt

tt

t

A=

++

+

=+

++

=+

++

+

=+

()(

)(

)()

()

102

22

1020

210

204

41

1 22

2

2

244

24

641

4

264

1 22

2

t

st

tt

tB

+

=+

+

=+

+

()

()


ex

am

ple 4

If A

star

ts 2

seco

nds b

efor

e B

find

whe

n an

d w

here

they

are

toge

ther

. Fin

d th

e m

axim

um d

ista

nce

that

A m

oves

ahe

ad o

f B in

the

subs

eque

nt m

otio

n.So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

S

st

tA=

++

=+

+=

++

=

2

2

1424

814

824

6411

224

200

()

()

m

st

tA=

++

=+

+=

++

=

2

2

1424

514

524

2570

2411

9

()

()

m

Dis

tanc

e af

ter 5

s:

Dis

tanc

e af

ter 8

s:

sut

at=

+1 2

2

s=

s=

119

mA

B

A

B

s=

s=

20

0 m

AB

u =

10

m s

a =

2 m

s

-1

-2

u =

1 m

sa

= 4

m s-1

-2

64

mt

t +

2

t =

0 s

t =

5 s

t =

8 s

s=

t+

14t +

24

A

2

s=

2t

+ t +

64

B

2

(t +

2) =

2 s

(t +

2)=

7 s

(t +

2)=

10

s

Co

nt.

....

sepA

rAt

ion

eq

uAt

ion

: s

st

tA

B−

=−

+−

213

40


ex

am

ple 4

If A

star

ts 2

seco

nds b

efor

e B

find

whe

n an

d w

here

they

are

toge

ther

. Fin

d th

e m

axim

um d

ista

nce

that

A m

oves

ahe

ad o

f B in

the

subs

eque

nt m

otio

n.So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

S

The

max

imum

sepa

ratio

n oc

curs

whe

n th

e ve

loci

ties o

f A

and

B a

re e

qual

.

Max

imum

Sep

arat

ion:

vv

AB

=

sut

at=

+1 2

2

s=

s=

119

mA

B

A

B

s=

s=

20

0 m

AB

u =

10

m s

a =

2 m

s

-1

-2

u =

1 m

sa

= 4

m s-1

-2

64

mt

t +

2

t =

0 s

t =

5 s

t =

8 s

s=

t+

14t +

24

A

2

s=

2t

+ t +

64

B

2

(t +

2) =

2 s

(t +

2)=

7 s

(t +

2)=

10

s

sepA

rAt

ion

eq

uAt

ion

: s

st

tA

B−

=−

+−

213

40

vuat

tt

t

vuatt

t

vv

A B AB

=+

=+

+=

++

=+

=+

=+

=+

=

102

210

24

214

14

41

2

()(

)

()

ttt

tt

ss

tt

AB

+=

+=

∴= −

=−

+−

=−

+−

=

144

113

2 65

1340

65

136

540

2

2

.

(.

)(

.)

s 2225.

m


ExErcisE 4. catch up1. Two bodies start together at the same time at the same place and move along the same straight line. If one moves with a constant speed of 8 m s–1 while the other starts from rest and moves at a constant acceleration of 2 m s–2. How long will it take before they are together?

2. A car A passes a point P on a straight road at a constant speed of 10 m s–1. At the same time another car B starts from rest at P with uniform acceleration 2.5 m s–2. (i) When and how far from P will B overtake A. (ii) If B ceases to accelerate on overtaking, what time elapses between the two cars passing a point Q which is 3 km from P.

3. A boy runs at 4 m s–1 away from a cyclist who starts at rest and accelerates at 2 m s–2. If the boy has an initial lead of 5 m, how long does the cyclist take to catch him?

4. Two bodies A and B travel in the same direction along the same line. Body A starts with velocity 3 m s–1 and acceleration 2 m s–2. The other body starts from the same place with velocity 1 m s–1 and acceleration 3 m s–2. Find when and where they are together again.

5. Two bodies move along parallel tracks in the same direction. Body A starts with velocity 2 m s–1 and acceleration 6 m s–2. Body B starts from the same place and the same time with velocity 5 m s–1 and acceleration 2 m s–2. Find when and where they are together again. Find their velocities when they are together for the second time.

6. Two bodies move in the same direction along parallel paths. A starts from point O with velocity 2 m s–1 and acceleration 4 m s–2. B starts 6 m ahead of A with velocity 3 m s–1 and acceleration 2 m s–2. Find when and where they are together and their velocities at this instant.

7. Two bodies move in the same direction along parallel paths. A starts from a point O with velocity 8 m s–1 and acceleration 2 m s–2 and B starts 8 m ahead of A and moves off with velocity 2 m s–1 at acceleration 4 m s–2. Find when they will be together and their distances from O at these times.

8. Two bodies A and B travel in the same direction along the same line from the same point P at the same time. A starts with velocity 5 m s–1 and acceleration 3 m s–2. B starts with velocity 2 m s–1 and acceleration 4 m s–2. They are together again at point Q. Find the time at which they are together and the distance | PQ |. Find their maximum distance apart between P and Q.

9. Two bodies move in the same direction along parallel paths. They start at the same point P at the same time. A starts from P with velocity 3 m s–1 and acceleration 2 m s–2. B starts with velocity 1 m s–1 and acceleration 3 m s–2. They are together again at point Q. Find the time at which they are together and the distance | PQ |.Find their maximum distance apart between P and Q.

10. Two bodies A and B move along parallel straight lines in the same direction from the same point P. A starts with velocity 4 m s–1 and acceleration 2 m s–2. B starts 1 second after A with velocity 2 m s–1 and acceleration 4 m s–2. Find when and where they will be together.


11. Two bodies A and B move along parallel straight lines in the same direction from the same point P. A starts from point P with velocity 5 m s–1 and acceleration 4 m s–2. B starts 1 second before A with velocity 6 m s–1 and acceleration 3 m s–2 from a point a distance of 2.5 m to the right of P. Find when and where they are together.

12. Find when and where they are together. Find their maximum separation between the two times when they are together.

13. IfAstarts2secondsbeforeBfindwhenand where they are together. Find their maximum separation between the two positions.

14. A car A starts from a point P with initial velocity 8 m s–1 and then travels with uniform acceleration 4 m s–2. Two seconds later a second car B starts from P with an initial velocity of 30 m s–2 and then moves with a uniform acceleration of 3 m s–2. Show that after passing A, B will never be ahead by more than 74 m.

15. Bodies A and B start together and move along the same straight line. A starts with a speed of 10 m s–1 and moves with a constant deceleration, while B starts at 5 m s–1 and accelerates at 4 m s–2. Find the deceleration of A if they meet when the velocity of B is twice that of A.

16. The driver of a car travelling at 20 m s–1 sees a second car 120 m in front travelling in the same direction at a uniform speed of 8 m s–1. (a) What is the least uniform retardation that must be applied to the faster car to avoid collision? (b) If the actual retardation is 1 m s–2find (i) the time interval in seconds for the faster car to reach a point 66 m behind the slower car, (ii) the shortest distance between the cars.

A

B

64 m

A

B

8 m

u = 10 m s–1 a = 2 m s–2

u = 1 m s–1 a = 4 m s–2

u = 10 m s–1 a = 2 m s–2

u = 1 m s–1 a = 4 m s–2


answErs

ExErcisE 4

1. 8 s

2. (i) 8 s, 80 m (ii) 146 s

3. 5 s

4. 4 s, 28 m from starting point

5. 1.5 s, 9.75 m, 11 m s–1, 8 m s–1

6. 3 s, 24 m, 14 m s–1, 9 m s–1

7. 2 s, 4 s, 20 m, 48 m

8. 6 s, 84 m, 4.5 m

9. 4 s, 28 m, 2 m

10. 5 s after B starts, 60 m

11. 10 s after A starts, 250 m from P

12. 1 s, 8 s, 12.25 m

13. 5 s and 8 s after B starts, 119 m from A and 200 m from A, 2.25 m

15. 4 m s–2

16. (a) 0.6 m s–2 (b) (i) 6 s, 18 s; (ii) 48 m


Sec

tio

n 4

: Ve

lo

cit

y-t

ime G

ra

phS

ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly

dece

lera

tes u

nifo

rmly

to re

st fo

r 4 s.

Dra

w a

vel

ocity

-tim

e cu

rve

to re

p-re

sent

the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.

ex

am

ple 2

A c

ar tr

avel

s fro

m A

to B

. It s

tarts

from

rest

at A

and

acc

eler

ates

at

2 m

s–2 u

ntil

it re

ache

s a sp

eed

of 3

0 m

s–1. I

t the

n tra

vels

at t

his s

peed

fo

r 600

m a

nd th

en d

ecel

erat

es a

t 2.5

m s–2

to c

ome

to re

st a

t B. F

ind

(i)

the

tota

l tim

e fo

r the

jour

ney,

(ii

) th

e di

stan

ce fr

om A

to B

,(ii

i)

the

aver

age

spee

d fo

r the

jour

ney.

ex

am

ple 3

A p

artic

le P

with

spee

d 14

0 m

s–1 b

egin

s to

dece

lera

te u

nifo

rmly

at a

ce

rtain

inst

ant w

hile

ano

ther

par

ticle

Q st

arts

from

rest

6 s

late

r and

ac-

cele

rate

s uni

form

ly. W

hen

the

seco

nd p

artic

le Q

has

trav

elle

d 12

5 m

, bo

th p

artic

les h

ave

a sp

eed

of 2

5 m

s–1.

(i)

Show

the

mot

ion

of b

oth

on th

e sa

me

spee

d-tim

e cu

rve.

(ii)

How

man

y se

cond

s afte

r the

com

men

cem

ent o

f dec

eler

atio

n do

es

th

e fir

st p

artic

le P

com

e to

rest

?


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

8 6 416 14 12 10

20

24

28

32

48

1216

v (m

s)

-1

t (s

)

2 0

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n =

Slop

e of

cur

veD

ista

nce

= A

rea

unde

r cur

ve

co

nt...

.


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

S

8

Run =

8 s

Ba

se

b =

8

14 12 10

20

24

28

32

48

1216

v (m

s)

-1

t (s

)

16

a = 2 ms

-2

6 4 26

4 m

0

Ris

e =

16

He

ight h

= 1

6m

s-1

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve c

on

t...

.

a sbh

==

=

==

=

−−

Ris

eR

unm

s sm

s

m

168

2

816

64

12

1 21 2(

)()

Are

a of

a tr

iang

le =

1 2bh

(i)


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

8 6 4

Leng

th l

= 2

0

14 12 10

20

24

28

32

48

1216

v (m

s)

-1

t (s

)

216

a = 2 ms

-2

64

m

Bre

ad

th b

= 1

6

32

0 m

0

mat

he

mat

ica

l c

al

cu

lat

ion

S

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve c

on

t...

.

a sbh

==

=

==

=

−−

Ris

eR

unm

s sm

s

m

168

2

816

64

12

1 21 2(

)()

Are

a of

a re

ctan

gle

= lb

×

(i) (ii)slb

=×

==

()(

)20

1632

0m


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

8 6 4

Run =

4 s

Ba

se

b =

4

14 12 10

20

24

28

32

48

1216

v (m

s)

-1

216

He

ight h

= 1

6

a = -4 m s-2

64

m3

20

m

a = 2 ms

-2

t (s

)3

2 m

Ris

e =

16

m s

-1

0

mat

he

mat

ica

l c

al

cu

lat

ion

S

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve

co

nt...

.

a sbh

==

=

==

=

−−

Ris

eR

unm

s sm

s

m

168

2

816

64

12

1 21 2(

)()

(i) (ii)slb

=×

==

()(

)20

1632

0m

Are

a of

a tr

iang

le =

1 2bh

(iii)a s

bh

==−

=−

==

=−−

Ris

eR

unm

s sm

s

m

164

4

416

3211

1 21 2(

)()


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

mat

he

mat

ica

l c

al

cu

lat

ion

S

co

nt...

.

(iv)

8 6 414 12 10

20

24

28

32

48

1216

v (m

s)

-1

216

t (s

)

a = -4 m s-2

64

m +

32

0 m

+ 3

2 m

= 4

16 m

a = 2 ms

-2

0

Ave

rage

Vel

ocity

Tota

l Dis

tanc

eTo

tal T

ime

=

Aver

age

velo

city

==

−41

6 3213

1m s

ms


ex

am

ple 1

A c

ar st

artin

g fr

om re

st a

ccel

erat

es u

nifo

rmly

ove

r 8 s

to a

vel

ocity

of

16 m

s–1. I

t the

n m

aint

ains

a c

onst

ant v

eloc

ity fo

r the

nex

t 20

s. It

final

ly d

ecel

erat

es u

nifo

rmly

to re

st fo

r 4 s.

Dra

w

a ve

loci

ty-ti

me

curv

e to

repr

esen

t the

mot

ion

of th

e ca

r. U

se th

e gr

aph

to fi

nd(i)

th

e ac

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r in

the

first

8 s,

(ii)

the

dist

ance

trav

elle

d by

the

car o

ver t

he n

ext 2

0 s,

(iii)

th

e de

cele

ratio

n an

d di

stan

ce tr

avel

led

by th

e ca

r ove

r the

last

4 s,

(iv)

the

aver

age

velo

city

of t

he c

ar fo

r its

ent

ire jo

urne

y.So

lu

tio

n

8 6 414 12 10

20

24

28

32

48

1216

v (m

s)

-1

216

t (s

)

32

a = -4 m s-2

a = 2 ms

-2

20

16

0

416

m

mat

he

mat

ica

l c

al

cu

lat

ion

S

Are

a=

+1 2(

)xyh

Fin

din

g t

he

ar

ea u

nd

er t

he

cu

rve

in o

ne

go

:A

trap

eziu

m is

a fo

ur si

ded

shap

e w

here

two

of th

e si

des a

re p

aral

lel.

The

area

of a

trap

eziu

m is

hal

f the

su

m o

f the

par

alle

l sid

es b

y th

e pe

rpen

dicu

lar d

ista

nce

betw

een

them

.

s=

+

= = =

1 2 1 2

2032

1652

1652

841

6()(

)(

)()

()(

) m


ex

am

ple 2

A c

ar tr

avel

s fro

m A

to B

. It s

tarts

from

rest

at A

and

acc

eler

ates

at

2 m

s–2 u

ntil

it re

ache

s a sp

eed

of 3

0 m

s–1. I

t the

n tra

vels

at t

his

spee

d fo

r 600

m a

nd th

en d

ecel

erat

es a

t 2.5

m s–2

to c

ome

to re

st

at B

. Fin

d(i)

th

e to

tal t

ime

for t

he jo

urne

y,

(ii)

the

dist

ance

from

A to

B,

(iii)

the

aver

age

spee

d fo

r the

jour

ney.

Sol

ut

ion

Tota

l tim

e T

= 15

s +

20 s

+ 12

s =

47 s

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve

mat

he

mat

ica

l c

al

cu

lat

ion

S

(i)I:aV t

t

t

=⇒

= ==

11

1230 30 2

15s

vs t

=s V t

= = =

−

600

301

2

m ms

?

II:V

s tt

t

=⇒

= ==

22

23060

0

600

3020

s

III:

.

.

aV t

t

t

=⇒

=

==

33

325

30

30 25

12s c

on

t....

.


ex

am

ple 2

A c

ar tr

avel

s fro

m A

to B

. It s

tarts

from

rest

at A

and

acc

eler

ates

at

2 m

s–2 u

ntil

it re

ache

s a sp

eed

of 3

0 m

s–1. I

t the

n tra

vels

at t

his

spee

d fo

r 600

m a

nd th

en d

ecel

erat

es a

t 2.5

m s–2

to c

ome

to re

st a

t B

. Fin

d(i)

th

e to

tal t

ime

for t

he jo

urne

y,

(ii)

the

dist

ance

from

A to

B,

(iii)

the

aver

age

spee

d fo

r the

jour

ney.

Sol

ut

ion

mat

he

mat

ica

l c

al

cu

lat

ion

S

(ii)A

rea=

+1 2(

)xyh

S=

+=

1 220

4730

1005

()

m

Ave

rage

Vel

ocity

Tota

l Dis

tanc

eTo

tal T

ime

=

(iii)

==

−10

05 4721

41

m sm

s.

Aver

age

velo

city

Tota

l tim

e T

= 15

s +

20 s

+ 12

s =

47 s


ex

am

ple 3

A p

artic

le P

with

spee

d 14

0 m

s–1 b

egin

s to

dece

lera

te u

nifo

rmly

at a

cer

tain

inst

ant w

hile

ano

ther

par

ticle

Q st

arts

fr

om re

st 6

s la

ter a

nd a

ccel

erat

es u

nifo

rmly

. Whe

n th

e se

cond

par

ticle

Q h

as tr

avel

led

125

m, b

oth

parti

cles

hav

e a

spee

d of

25

m s–1

.(i)

Sh

ow th

e m

otio

n of

bot

h on

the

sam

e sp

eed-

time

curv

e.(ii

) H

ow m

any

seco

nds a

fter t

he c

omm

ence

men

t of d

ecel

erat

ion

does

the

firs

t par

ticle

P c

ome

to re

st?

Sol

ut

ion

mat

he

mat

ica

l c

al

cu

lat

ion

S

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve

(ii) P

artic

le Q

:

125

255

101 21 2=

=

∴=

tt

t

()

s

sbh

=1 2

s h bt

= = =

−

125

251

m ms


ex

am

ple 3

A p

artic

le P

with

spee

d 14

0 m

s–1 b

egin

s to

dece

lera

te u

nifo

rmly

at a

cer

tain

inst

ant w

hile

ano

ther

par

ticle

Q st

arts

fr

om re

st 6

s la

ter a

nd a

ccel

erat

es u

nifo

rmly

. Whe

n th

e se

cond

par

ticle

Q h

as tr

avel

led

125

m, b

oth

parti

cles

hav

e a

spee

d of

25

m s–1

.(i)

Sh

ow th

e m

otio

n of

bot

h on

the

sam

e sp

eed-

time

curv

e.(ii

) H

ow m

any

seco

nds a

fter t

he c

omm

ence

men

t of d

ecel

erat

ion

does

the

firs

t par

ticle

P c

ome

to re

st?

Sol

ut

ion

mat

he

mat

ica

l c

al

cu

lat

ion

S

(ii) P

artic

le P

:

Ve

lo

cit

y t

ime c

ur

Ve

S

Acc

eler

atio

n a

= Sl

ope

of c

urve

Dis

tanc

e s =

Are

a un

der c

urve

a=

=−

=−

−R

ise

Run

ms

115

167

182

.

u a v t

= =−

= =

−

−

−

140 718

0

1

2

1ms m

sm

s.

?

vuat

tt

t

=+

=+−

=

==

014

07

187

1814

014

07

1819

5

(.

).

..

s


ExErcisE 5. VElocity timE curVEs

1. A car is travelling at 72 km h–1 when the brakes are applied producing a retardation of 4 m s–2. How long does it take to stop?

2. An electric train starts from a station and reaches a speed of 14 m s–1 in 25 s with uniform acceleration.Sketchthevelocity-timegraph,andfindhowfarithasgonebythetimeit reaches this speed.

3. An aircraft can take off when it reaches a speed of 180 km h–1. If it attains this speed in 30 s with uniform acceleration what distance does it require for taking off?

4. An express train is travelling at 144 km h–1 when its brakes are applied. If these produce a retardation of 2 m s–2 how long will it take to stop and what distance will it cover in doing so?

5. A train starts from rest and attains a speed of 50 km h–1 in 4 minutes with uniform acceleration. It runs at that speed for 5 minutes and then slows down uniformly to rest in 2minutes.Drawthevelocity-timegraphandfindthetotaldistancetravelled.

6. Find from the velocity-time graph shown (i) theaccelerationduringthefirst4s, (ii) the retardation during the last 2 s, (iii) the total distance travelled.

7. A cyclist rides along a straight road from A to B. He starts from rest at A and accelerates uniformly to reach a speed of 10 m s–1 in 8 s. He maintains this speed for 30 s and then uniformly decelerates to rest at B. If the total time is 48 s, draw a velocity-time curve and fromitfind (i) the acceleration, (ii) the deceleration, (iii) the total distance travelled.

8. A car travels from A to B. It starts from rest at A and accelerates at 1.5 m s–2 until it reaches a speed of 30 m s–1. It then travels at this speed for 2 km and then decelerates at 2 m s–2 to come to rest at B. Find (i) the total time for the journey, (ii) the distance from A to B, (iii) the average speed for the journey.

9. A and B are two points on a straight road. A car travelling along the road passes A when t = 0 and maintains a constant speed until t=20s,andinthistimecoversfour-fifthsofthe distance from A to B. The car then decelerates uniformly to rest at B. Draw a velocity-time curveandfindthetimefromAtoB.

t (s)

v (m/s)

0 84 10

8


10. A tram travels along a straight track and starts from rest. It accelerates uniformly for 20 s and during this time it travels 160 m. It maintains a constant speed for a further 50 s and decelerates to rest in 8 s. Calculate (i) the acceleration, (ii) the deceleration, (iii) the total time, (iv) the total distance.

11. A train starts from rest and travels 8 km in 12 minutes ending at rest. The acceleration is half the retardation, both are uniform, and there is a period when the train runs at its maximum speed of 50 km h–1. Find the time taken to reach full speed.

12. A 100 m sprinter starts with a speed of 6 m s-1 and accelerates uniformly to 10 m s–1 and finishestheraceatthisspeed.Ifhistotaltimeis10.4s,findhisuniformaccelerationand after what distance he is going at full speed.

13.Acartakes2minutestotravelbetweentwosetsoftrafficlights2145mapart.Ithasuniform acceleration for 30 s, then uniform velocity, and then uniform retardation for the last 15 s. Find the maximum velocity and the acceleration.

14. A train travels 15 km between two stations at an average speed of 50 km h–1. Its acceleration is half the retardation and both are uniform. If the maximum speed is 72 km h–1findthe acceleration in m s–2. Sketch the velocity-time curve.

15. A car accelerates at 2 m s–2 in bottom gear, 1.5 m s–2 in second gear and 1 m s–2 in top gear. Each gear change takes 1.5 s during which time the car travels at constant speed. If a motorist changes gear when his speeds are 3 m s–1 and 9 m s–1findhowlonghewilltaketo reach 15 m s–1 from rest.

16. A train moving in a straight line starts from A with uniform acceleration of 0.1 m s–2. After it has attained full speed it moves uniformly for 10 minutes. It is brought to rest at B by the brakes, which apply a constant retardation of 0.8 m s–2 for 20 s. Draw a rough velocity-time graphandfromitfindthetimeofthejourneyandthedistancefromA to B.

17. A train has a maximum speed of 72 km h–1 which it can achieve at an acceleration of 0.25 m s–2. With its brakes fully applied the train has a deceleration of 0.5 m s–2. What is the shortest time that the train can travel between stations 8 km apart if it stops at both stations?

18. A particle with speed 150 m s–1 begins to decelerate uniformly at a certain instant while another particle starts from rest 8 s later and accelerates uniformly. When the second particle has travelled 135 m, both particles have a speed of 30 m s–1. (i) Show the motion of both on the same speed-time curve. (ii)Howmanysecondsafterthecommencementofdecelerationdoesthefirstparticlecome to rest?


19. A body starts from rest at P travelling in a straight line and then comes to rest at Q which is 696 m from P.Thetimetakenis66s.Forthefirst10sithasuniformaccelerationa. It then travelsatconstantspeedandisfinallybroughttorestbyauniformdecelerationb acting for 6 s. Find a and b. If the journey from rest at P to rest at Q had been travelled with no interval of constant speed but at acceleration of a for a time t1 immediately followed by deceleration b for a time t2, show that the time for the journey is 8 29 s.

20. An athlete runs 100 m in 12 s. Starting from rest he accelerates uniformly to a speed of 10 m s–1 and then continues at that speed. Calculate the acceleration.

21. A cyclist has a maximum acceleration of 2 m s–2, a maximum speed of 15 m s–1 and a maximum deceleration of 4 m s–2. If he travels from rest to rest in the shortest possible time show that he covers a distance of 84 3

8 m. Find the time to travel (i) 105 m, (ii) 54 m.

AnswErs

ExErcisE 5

1. 5 s

2. 175 m

3. 0.75 km

4. 20 s, 400 m

5. 6 23 km

6. (i) 2 m s–2 (ii) 4 m s–2 (iii) 56 m

7. (i) 1.25 m s–2 (ii) 1 m s–2 (iii) 390 m

8. (i) 3053 s (ii) 2525 m (iii) 24.83 m s–1

9. 30 s

10. (i) 0.8 m s–2 (ii) 2 m s–2 (iii) 78 s (iv) 1024 m

11. 3.2 minutes

12. 2 m s–2, 16 m

13. 22 m s–1, 0.73 m s–2

14. 122

2ms−

15. 14.5 s

16. 13 minutes, 11.04 km

17. 460 s

18. (ii) 21.25 s

19. 1.2 m s–2, 2 m s–2

20. 2.5 m s–2

21. (i) 12 58 s (ii) 9 s


Sec

tio

n 5

: Fr

ee F

al

l

ex

am

ple 1

A b

ody

is th

row

n ve

rtica

lly u

p fr

om th

e gr

ound

at 1

4 m

s-1 .

Find

the

max

imum

hei

ght i

t rea

ches

and

the

time

to re

ach

this

hei

ght.

From

its h

ighe

st p

oint

find

the

time

for t

he b

ody

to h

it th

e gr

ound

and

its

spee

d w

hen

it hi

ts th

e gr

ound

.

ex

am

ple 2

A b

all i

s thr

own

verti

cally

up

at 1

0 m

s-1 f

rom

a p

oint

3 m

abo

ve th

e gr

ound

. Fin

d th

e sp

eed

it ha

s whe

n it

hits

the

grou

nd a

nd th

e tim

e it

take

s the

bal

l to

hit t

he g

roun

d.

ex

am

ple 3

A h

ot-a

ir ba

lloon

trav

els f

rom

the

grou

nd v

ertic

ally

up

at a

con

stan

t sp

eed

of 1

2 m

s-1 .

Find

its h

eigh

t abo

ve th

e gr

ound

afte

r 5 s.

A

fter 5

s a

ball

is d

ropp

ed fr

om th

e ba

loon

. How

long

doe

s it t

ake

to

reac

h th

e gr

ound

?


ex

am

ple 1

A b

ody

is th

row

n ve

rtica

lly u

p fr

om th

e gr

ound

at 1

4 m

s-1 .

Find

the

max

imum

hei

ght i

t rea

ches

and

the

time

to re

ach

this

hei

ght.

From

its h

ighe

st p

oint

find

its s

peed

whe

n it

hits

the

grou

nd a

nd th

e tim

e fo

r the

bod

y to

hit

the

grou

nd.

Sol

ut

ion

v=

0 m

s�1

s=

?

u=

14 m

s�1

GO

IN

GU

P

The

acce

lera

tion

due

to g

ravi

ty is

den

oted

by

g.Th

is v

alue

is 9

.8 m

s-2 .

In fr

ee fa

ll al

way

let a

= -

g = -9

.8 m

s-2 .

Go

inG

up:

u =

14 m

s-1

v =

0 m

s-1

a = -9

.8 m

s-1

s = ?

t = ?

vu

ass

s

s22

22

20

142

98

196

196

196

196

10

=+

=+

−=

∴=

=(.

).

.m

vuat

tt

t

=+

=−

=

==

=

014

98

98

1414 9

810 7

143

..

..

s

Go

inG

do

wn

:u

= 0

m s-

1

a = -9

.8 m

s-1

s = -

10 m

t = ?

v =

?

vu

asv v

v22

22

2

1

20

29

810

196 19

614

=+

=+

−−

=

∴=

=−

−

(.

)()

ms

vuat

t

t

=+

−=

−

==

=

140

98

14 98

10 71

43

.

..

s

Up

is p

ositi

veD

own

is n

egat

ive

Up

is p

ositi

veD

own

is n

egat

ive

u=

0 m

s�1

s=

10 m

�

v=

?

GO

IN

GD

OW

N

co

nc

lu

Sio

nS:

Tim

e up

= T

ime

Dow

n

Velo

city

goi

ng u

p =

Velo

city

at s

ame

poin

t on

way

dow

n bu

t in

the

oppo

site

dire

ctio

n


ex

am

ple 2

A b

all i

s thr

own

verti

cally

up

at 1

0 m

s-1 f

rom

a p

oint

3 m

abo

ve th

e gr

ound

. Fin

d th

e sp

eed

it ha

s whe

n it

hits

the

grou

nd a

nd th

e tim

e it

take

s the

bal

l to

hit t

he g

roun

d.So

lu

tio

n

u=

10 m

s�1

s=

3 m

�

v=

?

u =

10 m

s-1

s = -

3 m

a = -9

.8 m

s-1

v =

?t =

?

vu

asv v2

2

22 2

1

210

29

83

102

98

312

6

=+

=+

−−

=+

−−

=−

−

(.

)()

(.

)()

.m

s

vuat

tt

t

=+

−=

−−

=−

==

126

109

822

69

822

69

82

31

..

..

. ..

sTh

e ac

cele

ratio

n du

e to

gra

vity

is d

enot

ed b

y g.

This

val

ue is

9.8

m s-

2 .In

free

fall

alw

ay le

t a =

-g

= -9

.8 m

s-2 .

Up

is p

ositi

veD

own

is n

egat

ive

The

velo

city

is n

egat

ive

as it

is m

ovin

g do

wn.


ex

am

ple 3

A h

ot-a

ir ba

lloon

trav

els f

rom

the

grou

nd v

ertic

ally

up

at a

con

stan

t spe

ed o

f 12

m s-

1 . Fi

nd it

s hei

ght a

bove

the

grou

nd a

fter 5

s.

Afte

r 5 s

a ba

ll is

dro

pped

from

the

ballo

on. H

ow lo

ng d

oes i

t tak

e to

reac

h th

e gr

ound

? G

ive

your

ans

wer

to o

ne

plac

e of

dec

imal

.So

lu

tio

n

vs t

=v t vs t

svt

= = =⇒

=×

=×

=

−12 5

125

60

1m

ss

m

Bal

l is d

ropp

ed fr

om b

allo

on:

u =

12 m

s-1

s = -

60 m

a = -9

.8 m

s-2

t = ?

u=

12 m

s�1

t=

5 s

s=

?v

= 1

2 m

s�1

v=

12 m

s�1

Bal

l is

dro

pped

s=

60 m

�

t=

?

sut

at tt

tt

tt

t=+

−=

+−

−=

−

−−

=

∴=

1 22 1 2

2

2

2

6012

98

6012

49

49

1260

0

(.

)

..

11212

44

960

24

94

92

±−

−−

=(

)(

.)(

)(

.)

.s


exerCise 6. Uniformly g ACCelerAted motion

[In all problems g = 9.8 m s–2]

1. A vase falls from a shelf 140 cm above the floor. Find the speed with which it strikes the floor.

2. A stone is dropped from a point 49 m above the ground. Find the time for it to reach the ground.

3. A stone is thrown down at 5 m s–1. If its speed on hitting the ground is 19 m s–1 from what height was it thrown. How long does it take?

4. A stone is dropped from the top of a tower and falls to the ground. If it strikes the ground at 14 m s–1, how high is the tower?

5. A ball is thrown vertically downwards from the top of a tower with an initial speed of 2 m s–1. If it hits the ground 3 s later find (i) the height of the tower, (ii) the speed with which it hits the ground.

6. A stone is thrown upwards with a speed of 21 m s–1. Find its height (i) 1 s after projection, (ii) 2 s after projection, (iii) 3 s after projection.

7. A ball is thrown up at 14 m s–1 from a point 2 m above the ground. Find (i) the speed when it returns to the level of projection, (ii) the speed on the ground.

8. A ball is thrown vertically up at 28 m s–1. Find (i) the maximum height, (ii) the time to reach the maximum height, (iii) the velocity of return, (iv) the total time for the journey.

9. A balloon is rising at a steady speed of 3 m s–1. How high is it above the ground after 10 s? At this instant a man releases a stone. What is the initial velocity of the stone? How long does it take to reach the ground? How high is the balloon above the ground when the stone strikes the ground?

10. A stone is thrown up at 49 m s–1 from the ground. Find the times at which the particle is 78.4 m above the ground. Find the time interval for which the particle is above 78.4 m.

11. A ball is thrown up at 14 m s–1. Find the times at which the particle is 9.1 m above the ground.


12. A ball is thrown up at 49 m s–1. How long does it take to reach its maximum height? If an other ball was thrown up 1 s after the first one, how high is it above the ground when the first ball has reached its maximum height if it has the same initial velocity?

13. A jumper can jump 2 m on the Earth. What is his take-off speed? How high can he jump on the moon? (Acceleration due to gravity of moon g = 1.6 m s–1 )

14. A particle is thrown vertically upwards under gravity with a speed of 16 m s–1. One second later another particle is fired upwards from the same point. Find the initial speed of this particle in order that the two particles will collide when the first particle has reached its highest point.

15. An object falls vertically past a window 2 m high in 112 s. Find the height above the window

from which the object was dropped.

16. A stone is dropped from a balloon rising at 10 m s–1 and reaches the ground in 8 s. How high was the balloon above the ground when the stone was dropped?

17. A body falls from the top of a tower and during the last second it falls 925 of the total

distance. Find the height of the tower.

18. A particle falls freely from rest from a point O passing three points A, B and C, the distances |AB| and |BC| being equal. If the particle takes 3 s to pass from A to B and 2 s from B to C, calculate |AB|.

19. A body falls freely from rest from a point O passing three points A, B and C, the distances |AB| and |BC| being equal. The time taken to go from A to B is 2 s and from B to C is 1 s. Find |AB|.

20. A particle falls freely under gravity from rest at a point P. After it has fallen for 1 s another particle is projected vertically downwards from P with speed 14.7 m s–1. Find the time and distance from P at which they collide.


Answers

exerCise 6

1. 5.24 m s–1

2. 3.16 s

3. 17.14 m, 1.43 s

4. 10 m

5. (i) 50.1 m (ii) 31.4 m s–1

6. (i) 16.1 m (ii) 22.4 m (iii) 18.9 m

7. (i) 14 m s–1 (ii) 15.34 m s–1

8. (i) 40 m (ii) 207 s (iii) 28 m s–1

(iv) 40

7 s

9. 30 m, 3 m s–1, 2.8 s, 38.4 m

10. 2 s, 8 s, 6 s

11. 1 s, 137 s

12. 5 s, 117.6 m

13. 6.26 m s–1, 12.25 m

14. 23.7 m s–1

15. 28.4 m

16. 233.6 m

17. 122.5 m

18. 147 m

19. 29.4 m

20. 2 s, 19.6 m


20131. (a) A ball is thrown vertically upwards with a speed of 44·1 m s−1. Calculate the time interval between the instants that the ball is 39·2 m above the point of projection.

(b) A lift ascends from rest with constant acceleration f until it reaches a speed v. It continues at this speed for t1 seconds and then decelerates uniformly to rest with deceleration f. The total distance ascended is d, and the total time taken is t seconds.

(i) Draw a speed-time graph for the motion of the lift.

(ii) Show that v f t t= −12 1( ).

(iii) Show that t t df1

2 4= − .

leAving Cert QUestions

20141. (a) Two cars, P and Q, travel with the same constant velocity 15 m s–1 along a straight level road. The front of car P is 24 m behind the rear of car Q. At a given instant both cars decelerate, P at 4 m s–2 and Q at 5 m s–2.

(i) Find, in terms of t, the distance between the cars t seconds later.

(ii) Find the distance between the cars when they are at rest.

20151. (a) A particle starts from rest and moves with constant acceleration. If the particle travels 39 m in the seventh second, find the distance travelled in the tenth second.

(b) A train of length 66·5 m is travelling with uniform acceleration 47

2m s− . It meets another train of length 91 m travelling on a parallel track in the opposite direction with uniform acceleration 8

72m s− .

Their speeds at this moment are 18 m s–1 and 24 m s–1 respectively.

(i) Find the time taken for the trains to pass each other.

(ii) Find the distance between the trains 1 second later.


20121. (a) A particle falls from rest from a point P. When it has fallen a distance 19·6 m a second particle is projected vertically downwards from P with initial velocity 39·2 m s–1. The particles collide at a distance d from P. Find the value of d.

(b) A car, starts from rest at A, and accelerates uniformly at 1 m s–2 along a straight level road towards B, where AB = 1914 m. When the car reaches its maximum speed of 32 m s–1, it continues at this speed for the rest of the journey. At the same time as the car starts from A a bus passes B travelling towards A with a constant speed of 36 m s–1. Twelve seconds later the bus starts to decelerate uniformly at 0·75 m s–2.

(i) The car and the bus meet after t seconds. Find the value of t.

(ii) Find the distance between the car and the bus after 48 seconds.

20111. (a) A particle is released from rest at A and falls vertically passing two points B and C.

It reaches B after t seconds and takes 27t

seconds to fall from B to C, a distance of 2.45 m.

Find the value of t.

(b) A car accelerates uniformly from rest to a speed v in t1 seconds. It continues at this constant speed for t seconds and then decelerates uniformly to rest in t2 seconds.

The average speed for the journey is 34v .

(i) Draw a speed-time graph for the motion of the car.

(ii) Find t1 + t2 in terms of t.

(iii) If a speed limit of 23v were to be applied, find in terms of t the least time the

journey would have taken, assuming the same acceleration and deceleration as in part (ii).

A

B

C


20101. (a) A car is travelling at a uniform speed of 14 ms–1 when the driver notices a traffic light turning red 98 m ahead.

Find the minimum constant deceleration required to stop the car at the traffic light, (i) if the driver immediately applies the brake (ii) if the driver hesitates for 1 second before applying the brake.

(b) A particle passes P with speed 20 ms–1 and moves in a straight line to Q with uniform acceleration. In the first second of its motion after passing P it travels 25 m. In the last 3 seconds of its motion before reaching Q it travels 13

20 of |PQ|.

Find the distance from P to Q.

20091. (a) A particle is projected vertically upwards from the point P. At the same instant a second particle is let fall vertically from Q. The particles meet at R after 2 seconds.

The particles have equal speeds when they meet at R.

Prove that |PR| = 3|RQ|.

(b) A train accelerates uniformly from rest to a speed v m/s with uniform acceleration f m/s2.

It then decelerates uniformly to rest with uniform retardation 2f m/s2. The total distance travelled is d metres. (i) Draw a speed-time graph for the motion of the train.

(ii) If the average speed of the train for the whole journey is d3

, find the value of f.

Q

R

P


20081. (a) A ball is thrown vertically upwards with an initial velocity of 39.2 m/s. Find (i) the time taken to reach the maximum height

(ii) the distance travelled in 5 seconds.

(b) Two particles P and Q, each having constant acceleration, are moving in the same direction along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s, respectively. Two minutes later Q passes P, and Q is then moving at 65.5 m/s. Find (i) the acceleration of P and the acceleration of Q

(ii) the speed of P when Q overtakes it

(iii) the distance P is ahead of Q when they are moving with equal speeds.

20071. (a) A particle is projected vertically downwards from the top of a tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower. During the third second of its motion the particle travels 29.9 metres.

Find (i) the value of u

(ii) the height of the tower.

(b) A train accelerates uniformly from rest to a speed v m/s. It continues at this speed for a period of time and then decelerates uniformly to rest. In travelling a total distance d metres the train accelerates through a distance pd metres and decelerates through a distance qd metres, where p < 1 and q < 1. (i) Draw a speed-time graph for the motion of the train.

(ii) If the average speed of the train for the whole journey is vp q b+ +

,

find the value of b.

29.9 m

t = 0

t = 2

t = 3


20061. (a) A lift starts from rest. For the first part of its descent it travels with uniform acceleration f. It then travels with uniform retardation 3f and comes to rest. The total distance travelled is d and the total time taken is t. (i) Draw a speed-time graph for the motion.

(ii) Find d in terms of f and t.

(b) Two trains P and Q, each of length 79.5 m, moving in opposite directions along parallel lines, meet at O, when their speeds are 15 m/s and 10 m/s respectively. The acceleration of P is 0.3 m/s2 and the acceleration of Q is 0.2 m/s2. It takes the trains t seconds to pass each other. (i) Find the distance travelled by each train in t seconds.

(ii) Hence, or otherwise, calculate the value of t.

(iii) How long does it take for 25 of the length of train Q to pass the point O?

20051. (a) Car A and car B travel in the same direction along a horizontal straight road. Each car is travelling at a uniform speed of 20 m/s. Car A is at a distance of d metres in front of car B. At a certain instant car A starts to brake with a constant retardation of 6 m/s2. 0.5 s later car B starts to brake with a constant retardation of 3 m/s2 . Find (i) the distance travelled by car A before it comes to rest

(ii) the minimum value of d for car B not to collide with car A.

20041. (a) A ball is thrown vertically upwards with an initial velocity of 20 m/s. One second later, another ball is thrown vertically upwards from the same point with an initial velocity of u m/s. The balls collide after a further 2 seconds. (i) Show that u = 17.75.

(ii) Find the distance travelled by each ball before the collision, giving your answers correct to the nearest metre.


20031. (a) The points P, Q and R all lie in a straight line. A train passes point P with speed u m/s. The train is travelling with uniform retardation f m/s2. The train takes 10 seconds to travel from P to Q and 15 seconds to travel from Q to R, where | PQ | = | QR | = 125 metres.

(i) Show that f = 13 .

(ii) The train comes to rest s metres after passing R. Find s, giving your answer correct to the nearest metre.

(b) A man runs at constant speed to catch a bus. At the instant the man is 40 metres from the bus, it begins to accelerate uniformly from rest away from him. The man just catches the bus 20 seconds later. (i) Find the constant speed of the man.

(ii) If the constant speed of the man had instead been 3 m/s, show that the closest he gets to the bus is 17.5 metres.

20021. (a) A stone is thrown vertically upwards under gravity with a speed of u m/s from a point 30 metres above the horizontal ground. The stone hits the ground 5 seconds later. (i) Find the value of u.

(ii) Find the speed with which the stone hits the ground.

(b) A particle, with initial speed u, moves in a straight line with constant acceleration. During the time interval from 0 to t, the particle travels a distance p. During the time interval from t to 2t, the particle travels a distance q. During the time interval from 2t to 3t, the particle travels a distance r. (i) Show that 2q = p + r.

(ii) Show that the particle travels a further distance 2r – q in the time interval from 3t to 4t.


20011. (a) Points P and Q lie in a straight line, where | PQ | = 1200 metres. Starting from rest at P, a train accelerates at 1 m/s2 until it reaches the speed limit of 20 m/s. It continues at this speed of 20 m/s and then decelerates at 2 m/s2, coming to rest at Q. Find the time it takes the train to go from P to Q. Find the shortest time it takes the train to go from rest at P to rest at Q if there is no speed limit, assuming that the acceleration and deceleration remain unchanged at 1 m/s2

and 2 m/s2, respectively.

(b) A particle is projected vertically upwards with an initial velocity of u m/s and another particle is projected vertically upwards from the same point and with the same initial velocity T seconds later. Show that the particles

(i) will meet T u

g2+

seconds from the instant of projection of the first particle

(ii) will meet at a height of 4

8

2 2 2u g Tg-

metres.

20001. (a) A stone projected vertically upwards with an initial speed of u m/s rises 70 m in the first t seconds and another 50 m in the next t seconds. Find the value of u.

(b) A car, starting from rest and travelling from P to Q on a straight level road, where | PQ | = 10 000 m, reaches its maximum speed 25 m/s by constant acceleration in the first 500 m and continues at this maximum speed for the rest of the journey. A second car, starting from rest and travelling from Q to P, reaches the same maximum speed by constant acceleration in the first 250 m and continues at this maximum speed for the rest of the journey. (i) If the two cars start at the same time, after how many seconds do the two cars meet? Find, also, the distance travelled by each car in that time.

(ii) If the start of one car is delayed so that they meet each other exactly halfway between P and Q, find which car is delayed and by how many seconds.

19991 (b) A particle travels in a straight line with constant acceleration f for 2t seconds and covers 15 metres. The particle then travels a further 55 metres at constant speed in 5t seconds. Finally the particle is brought to rest by a constant retardation 3f.

(i) Draw a speed-time graph for the motion of the particle.

(ii) Find the initial velocity of the particle in terms of t.

(iii) Find the total distance travelled in metres, correct to two decimal places.


19981 (a) A train accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a period of time and then decelerates uniformly to rest. If the average speed for the whole journey is 5

6 v, find what fraction of the whole distance is described at constant speed.

(b) Car A, moving with uniform acceleration 320 b m/s2 passes a point P with speed 9u m/s.

Three seconds later car B, moving with uniform acceleration 29 b m/s2 passes the same

point with speed 5u m/s. B overtakes A when their speeds are 6.5 m/s and 5.4 m/s respectively. Find (i) the value u and the value b,

(ii) the distance travelled from P until overtaking occurs.

19971 (a) A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance travelled while accelerating is 6 m. The total distance travelled is 30 m and the total time taken is 6 s.

(i) Draw a speed-time graph and hence, or otherwise, find the value of v.

(ii) Calculate the distance travelled at v m/s.

(b) Two points P and Q are a distance d apart. A particle starts from P and move towards Q with initial velocity 2u and uniform acceleration f. A second particle starts at the same time from Q and moves towards P with initial velocity 3u and uniform deceleration f. Prove that

(i) the particles collide after du5

seconds,

(ii) if the particles collide before the second particle comes to instantaneous rest, then

fd u<15 2,

(iii) if fd u= 30 2 then the second particle has returned to Q before the collision.


19961 (a) A particle starts from rest and moves in a straight line with uniform acceleration. It passes three points A, B and C where |AB| = 105 m and |BC| = 63 m. If it takes 6 seconds to travel from A to B and 2 seconds to travel from B to C find (i) its acceleration

(ii) the distance of A from the starting position.

(b) A lift starts from rest with constant acceleration 4 m/s2. It then travels with uniform speed and finally comes to rest with constant retardation 4 m/s2. The total distance travelled is d and the total time taken is t. (i) Draw a speed-time graph.

(ii) Show that the time for which it travelled with uniform speed is t d2 - .

Answers

leAving Cert. QUestions

2015. 1 (a) 57 m (b) (i) 3·5 s, (ii) 48·9 m

2014. 1 (a) (i) 24 12

2- t , (ii) 18.375 m

2013. 1 (a) 7 s

2012. 1 (a) 44.1 m (b) (i) 40 s, (ii) 352 m

2011. 1 (a) t = 78 s

(b) (ii) t = t1 + t2, (iii) 3112 t

2010. 1 (a) (i) –1 m s–2, (ii) - -76

2m s (b) 300 m

2009. 1 (b) (ii) f = 1 m s–2

2008. 1 (a) (i) 4 s, (ii) 83.3 m (b) (i) a aP Q= =5

242 20 5m/s m/s, . (ii) 48 m/s (iii) 525 m

2007. 1 (a) (i) 5.4 m/s, (ii) 100 m; (b) (ii) b = 1

2006. 1 (a) (ii) 38

2ft ;

(b) (i) s t t s t tP Q= + = +15 0 15 10 0 12 2. , . , (ii) 6 s, (iii) 3.1 s

2005. 1 (a) (i) 1003 m, (ii) 43.3 m

2004. 1 (a) (ii) 25 m, 16 m


2003. 1 (a) (ii) 51 m; (b) (i) 4 m/s

2002. 1 (a) (i) 18.5 m/s, (ii) 30.5 m/s;

2001. 1 (a) 75 s, 60 s;

2000. 1 (a) 56 m/s; (b) (i) 215 s, P: 4875 m, Q: 5125 m, (ii) Delay car starting at Q by 5 s

1999. 1 (b) (ii) 4t

, (iii) 75.76 m

1998. 1 (a) 45

(b) (i) u = 0.1 m/s, b = 1, (ii) 94.5 m

1997. 1 (a) (i) 6.5 m/s, (ii) 21 m;

1996. 1 (a) (i) 3.5 m/s2, (ii) 7 m


6th Year Applied Maths Higher Level Kieran Mills Uniform ...

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