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Motion in a Straight Line

Jr|11th|I PUC IIT-JEE ADVANCED|PHYSICS:VOL-I

3

A. Motion along Horizontal Axis :SINGLEANSWER TYPE

1. A body moving along a straight line traversedone third of the total distance with a velocity 4m/sec in the first stretch. In the second stretchthe remaining distance is covered with avelocity 2 m/sec for some time 0t and with 4m/s for the remaining time. if the average velocityis 3 m/sec, find the time for which body moveswith velocity 4 m/sec in second stretch:

A) 0

3t

2B) t

0C) 2t

0D) 0t

22. The acceleration of the particle is increasing

linerly with time ‘t’ as bt. The particle starts fromthe origin with an initial velocity v0. The distancetravelled by the particle in time ‘t’ will be

A) 3

0

1

6v t bt+ B)

30

1

3v t bt+

C) 2

0

1

3v t bt+ D)

20

1

2v t bt+

3. A particle moves along the positive branch

of the curve 2

2

xy = with x governed by

2

2

tx = where x and y are measured in meters

and t in seconds. At t=2s, the velocity of theparticle is

A) �2 3 /i jm s−� B) �2 4 /i jm s+�

C) �4 2 /i jm s+� D) �4 2 /i jm s−�

4. A point moves in a straight line so that its

displacement x metre time t second is given

by 2 21x t= + . Its acceleration in 2ms− at timet second is

A) 3/ 2

1

x B) 3

tx−

C) 2

3

1 tx x

− D) 2

1 1

x x−

5. A 2m wide truck is moving with a uniformspeed 1

0 8v ms−= along a straight horizontalroad. A pedestrain strarts to cross the roadwith a uniform speed v when the truck is 4 maway from him. The minimum value of v sothat he can cross the road safely is

A) 12.62 ms− B) 14.6 ms−

C) 13.57 ms− D) 11.414 ms−

6. In 1.0 s, a particle goes from point A to pointB, moving in a semicircle (see figure). Themagnitude of the average velocity is

A) 3.14 m/s

1.0m

A

B

B) 2.0 m/s

C) 1.0 m/s

D) zero

3

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4Motion in a Straight Line


MULTIPLE ANSWER QUESTIONS7. The velocity of a particle along a straight line

increases according to the linear law v =v0+kx, where k is a constant. ThenA) the acceleration of the particle is ( )0k v kx+

B) the particle takes a time 1e

0

v1log

k v

⎛ ⎞⎜ ⎟⎝ ⎠

to attain a

velocity v1

C) velocity varies linearly with displacement withslope of velocity displacement curve equal to k.D) data is insufficient to arrive at a conclusion.

8. Two particles P and Q move in a straight lineAB towards each other. P starts from A withvelocity u1 and an acceleration a1. Q starts fromB with velocity u2 and acceleration a2. They passeach other at the midpoint of AB and arrive atthe other ends of AB with equal velocities.

A) They meet at midpoint at time ( )( )

2 1

1 2

2 u ut

a a

−=

−B) The length of path specified i.e., AB is

( )( )( )1

2 1 1 2 2 12

2

4 u u a u a ul

a a

− −=

−

C) They reach the other ends of AB with equal

velocities if ( )( ) ( )2 1 1 2 1 2 2 1u u a a 8 a u a u+ − = −D) They reach the other ends of AB with equal

velocities if ( )( ) ( )2 1 1 2 2 1 1 2u u a a 8 a u a u− + = −9. Which of the following statements is/ are correct ?

A) If the velocity of a body changes, it must havesome acceleration.B) If the speed of a body changes, it must havesome accelerationC) If the body has acceleration, its speed mustchangeD) If the body has acceleration, its speed maychange.

10. A particle moves along a straight line so thatits velocity depends on time as 24v t t= − .Then for first 5s.A) Average velocity is 125 / 3 ms−

B) Average speed is 110 ms−

C) Average velocity is 15 / 3 ms−

D) Acceleration is 24 0ms at t− =11. A particle moves with an initial velocity v0 and

retardation vα , where v is velocity at any time t.

A) The particle will cover a total distance 0vα

B) The particle will come to rest after time 1

αC) The particle will continue to move for a longtime.

D) The velocity of particle will become 0ve after

time 1

α12. A particle is moving along X–axis whose

position is given by 3t

x 4 9t3

= − + . Mark the

correct statement(s) in relation to its motion.A) direction of motion is not changing at any of theinstantsB) direction of motion is changing at t = 3 sC) for 0 < t < 3 s, the particle is slowing downD) for 0 < t < 3 s, the particle is speeding up.

13. A particle of mass m moves on the x-axis asfollows : it starts from rest at t = 0 from thepoint x = 0, and comes to rest at t = 1 at thepoint x = 1. No other information is availableabout its motion at intermediate times( )0 t 1< < . If a denotes the instantaneousacceleration of the particle, then [1993]A) a cannot remain positive for all t in the interval0 t 1≤ ≤B) |a| cannot exceed 2 at any point in its pathC) |a| must be ≥ 4 at some point or points in itspathD) a must change sing during the motion, but noother assertion can be made with the informationgiven.

PASSAGE TYPE QUESTIONSPassage-1

A train starts from rest with constantacceleration, 2a 1m / s= . A passenger at adistance ‘S’ from the train runs at hismaximum velocity of 10 m/s to catch the trainat the same moment at which the train starts.

14. If S=25.5 m and passenger keeps running, findthe time in which he will catch the train:A) 5 sec B) 4 sec C) 3 sec D) 2 2 sec.

15. Find the critical distance ‘Sc’ for whcihpassenger will take the ten seconds time tocatch the train:A) 50m B) 35m C) 30m D) 25m

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16. Find the speed of the train when the passengercatches it for the critical distacne:A) 8 m/s B) 10 m/s C) 12 m/s D) 15m/s

Passage-2A body is moving with uniform velocity of

18 ms− . When the body just crossed anotherbody, the second one starts and moves withuniform acceleration of 24 ms− .

17. The time after which two bodies meet will beA) 2 s B) 4 s C) 6 s D) 8 s

18. The distance covered by the second bodywhen they meet isA) 8 m B) 16 m C) 24 m D) 32 m

MARTIX MATCHING QUESTION19. A particle moves along a straight line such

that its displacement S varies with timeas 2S t tα β γ= + + .Column-Ii. Acceleration at t= 2 s

ii. Average velocity during 3rd sec

iii. Velocity at 1t s=iv. Initial displacementColumn-IIa. 5β γ+ b. 2γc. α d. 2β γ+

INTEGER TYPE QUESTIONS20. In a car race, car A takes 4 s less than car B at

the finish and passes the finishing point with avelocity v more than the car B. Assuming thatthe cars start from rest and travel with constantaccelerations 2

1 4a ms−= and 22 1a ms−=

respectively, find the velocity of v in 1ms− .21. A police jeep is chasing a culprit going on a

motorbike. The motorbike crosses a turningat a speed of 172 kmh− . The jeep follows it ata speed of 190 kmh− , crossing the turning 10s later than the bike.Assuming that they travel at constant speeds,how far from the turning will the jeep catchup with the bike ? ( in km)

22. A particle moves in a straight line such thatthe displacement x at any time ‘t’ is given by

2 36 3 4.x t t t= − − − x is in m and t is in secondcalculate the maximum velocity (In ms-1) ofthe particle.

B. Motion Under GravitySINGLE ANSWER QUESTIONS

23. A ball is thrown upwards with speed v from thetop of a tower and it reaches the ground withspeed 3v . What is the height of the tower ?

A) 2v

g B) 22v

g C) 24v

g D) 28v

g24. An elevator in which a man is standing is

moving upwards with a speed of 110 ms− . Ifthe man drops a coin from a height of 2.45 mfrom the floor of elavator, it reaches the floorof the elavator after time ( 29.8g ms−= )

A) 2 s B) 1/ 2 s C) 2 s D) 1/ 2 s25. A body is thrown vertically upwards from A, the

top of a tower. It reaches the ground in time 1t .If it is thrown vertically downward from A withthe same speed, it reaches the ground in time

2t . If it is allowed to fall freely from A, then thetime it takes to reach the ground is given by

A) 1 2

2

t tt += B) 1 2

2

t tt −=

C) 1 2t t t= D) 1

2

ttt

=

26. A stone is dropped from the 25th storey of amultistoried building and it reaches theground in 5s. In the first second, it passesthrough how many storeys of the building

( g= 210 ms− )A)1 B)2 C) 3 D) None

27. A body thrown up in a lift with an upwardvelocity u relative to the lift from its floorand the time of flight is found to be ‘t’. Theacceleration of the lift will be

A) 2

u gt−B)

2

u gt+C)

2u gtt−

D) u gt

−

28. A ball is dropped vertically from a height dabove the ground. It hits the ground and

bounces up vertically to a height 2

d .

Neglecting subsequent motion and airresistance, its velocity v varies with height habove the ground as: [2004]

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A)

V

dh

B)

V

d h

C)

V

dh

D)

V

dh

29. A small block slides without friction down aninclined plane starting from rest. Let sn be thedistance traveled from t =n – 1 to t = n. Then

1

n

n

ss +

is : [2005]

A) 2 – 1

2

nn B)

2 1

2 – 1

nn

+C)

2 – 1

2 1

nn + D)

2

2 1

nn +

MULTIPLE ANSWER QUESTIONS30. S

1,S

2 and S

3 are the different sizes of windows

1,2 and 3 respectively, placed in avertical plane.A particle is thrown up in that vertical plane. Findthe correct options:

S3

S2

S1

A) average speed of the particlepassing the windows may be equal if

1 2 3s s s< <B) average speed of the particlepassing the windows may be equal if

1 2 3S S S> >

C) If 1 2 3S S S= = , the change inspeed of the part icle whilecrossing the windows will satisfy

1 2 3V V V∆ < ∆ < ∆ .

D) If 1 2 3S S S= = , the time taken byparticle to cross the windows will

satisfy 1 2 3t t t< < .

31. At 0t = , an bullet is fired vertically upwardwith a speed of 100ms–1. A second bullet isfired vertically upwards with the same speedat 5 .t s= ThenA) The two bullets will be at the same heightabove the ground at 12.5t s=B) The two bullets will reach back their strartingpoints at 20t s= and at 25t s=

C) The ratio of the speeds of the first and secondbulletsat 20t s= will be 2: 1D) The maximum height attained by either bulletwill be 1000 m

32. From the top of a tower of height 200 m, aball A is projected up with 110 ms− , and 2 slater another ball B is projected verticallydown with the same speed ThenA) Both A and B will reach the groundsimultaneouslyB) Ball A will hit the ground 2 s later than B hittingthe ground.C) Both the balls will hit the ground with the samevelocityD) Both the balls will hit the ground with thedifferent velocity

MATRIX MATCH QUESTIONS33. A particle moves such that , t x 3= + , where

x is in metre, t is in second. Based on thisinformation, match the value in Column-I (inSI units) to their respective quantities for theparticles motion given in Column-IIColumn-I Column-II

A) 0 p) Acceleration at t= 5 sB) 2 q) Average speed from t =0 to t =6sC) 3 r) Velocity at the point of reversal of

motionD) 18 s) Total distance travelled from t =0 to

t=6st) Displacement from t =0 to t=6s

34. For a body projected vertically up with a velocityv0 from the ground, match the followingColumn-IA. avV��

Average velocity)

B. avU ( Average speeD)

C. ascentTD. descentTColumn-IIi. Zero for round trip

ii. 1 2

2

v v+� �

over any time interval where 1 2v & v� �

are

the intial and final velocities in the time interval

iii. 0

2

v over the total timeof its flight

iv. 0v

g

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INTEGER TYPE QUESTIONS35. From a lift moving upwards with a uniform

acceleration 22 ,a ms−= a man throws a ball

vertically upwards with a velocity 112v ms−=relative to the lift. The ball comes back tothe man after a time .t Find the value of t insecond (g=10ms-2).

36. A body is thrown up with a velocity 100ms–1.It travels 5 m in the last second of upwordjourney. If the same body is thrown up with avelocity 200ms–1, how much distance ( inmetre ) will it travel in the last second of itsopward journey (g=10ms–2)

C. Graphs

SINGLE ANSWER QUESTIONS37. The following graph shows the variation of

velocity of a rocket with time. Then themaximum height attained by the rocket is

v(ms )–1

1000

0 10100

120t(s)

A) 1.1 km B) 5 km C) 55 km D) None38. The velocity-time graph of a particle moving

in a straight line is shown in figure. Theacceleration of the particle at 9t = is

2 4 6 8 10 12

t(s)

v(ms )–1

15

10

5

A) Zero B) 25 ms− C) 25 ms−− D) 22 ms−−

MULTIPLE ANSWER QUESTIONS39. Figure shows the velocity( v) of a particle

plotted against time ( t).

vT

O2T

t

A) The particle changes its direction of motion atsome point

B) The acceleration of the particle remainsconstant .C) The displacement of the particle is zeroD) The initial and final speeds of the particle arethe same.

40. A particle moves in astraight with thevelocity as shown in figure. At

0, 16t x m= = − .

t(s)4030241810

6

2

0–2

–6

V

A) The maximum value of the position coordinateof the particle is 54 m.B) The maximum value of the position coordinateof the particle is 36 m.C) The particle is at the position of 36 m at

18t s= .D) The particle is at the position of 36 m at

30t s=

NUMERICAL GRADING41. A proton moves along x-axis according to

the equation 50

10

x tt −= where x is an

metres and ‘t’ is in seconds. Calculate theaverage velocity in ms–1 of the proton duringthe first 3.0 sec of its motion?

42. The buses A and B are at positions 50m and100m from the origin at time t=0. They startmoving in the same direction simultaneouslywith uniform velocity of 10ms–1 and 5ms–1.Find the position in metres from the origin atwhich A overtakes B?

43. A stone is dropped into a well of depth 67.6m.The splash of water is heard after 3.91sec.Find the velocity of sound in ms–1?

44. A parachutist drops freely from an aeroplanefor 10 seconds before the parachute opensout. Then he descends with a net retardationof 2.5ms–2 and strikes the ground with avelocity 5ms–1. Find the height in metresfrom which he bails out of the plane? (Takeg=10ms-2)

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1) D 2) B 3) b 4) C 5) C 6) B7) A,B,C 8) A,B,C 9) A,B,D10) C,D 11) A,C,D 12) B,C13) A,C 14) C 15) A 16) B 17) B 18) D19) i - b, ii - a, iii - d, iv - c20) 8 21) 1 22) 9 23) C 24) B 25) C26) A 27) C 28) A 29) C 30) B,C,D31) A,B,C32) A,C33) a- (r,t) b-(p) c-(q) d-(s)34) A - i,ii B - iii, C - iv, D - iv35) 2 36) 5 37) C 38) C 39) A,B,C,D40) A,C,D41) 0080 42) 0150 43) 033844) 2495

1. Sol: 1

s / 3 st

4 12= =

( ) ( ) ( )0 0 0

2s2 t 4 kt t 2 4k

3= + = + or ( )0

2st

3 2 4k=

+

Average velocity = 1 0 0

s

t t t k+ +

( )( )

( )( )

6 2 4ks2s 1 k 5 6ks

12 3 2 4k

+= =

+ ++

+

( )avv 5 6k 12 24k+ = + gives 1

k2

=

Required time = 00

tkt

2= .

2. Given, acceleration a=bt2

2

dv btbt v cdt

⇒ = ⇒ = +

At 0 00,t v v c v= = ⇒ =

So, 2

02

btv v= +

2

02

ds bt vdt

⇒ = +3

06

bts v t⇒ = +

3.2

2 xt dxx v t

dt= ⇒ = = and

( )2 22 4

2

2 2 8

tx ty⇒ = = =

3

2ydy tvdt

= =

4. 2 21x t= + or ( )1/ 221x t= +

( ) ( )1/ 2 1/ 22 211 2 1

2

dx t t t tdt

− −= + = +

( ) ( )2

1/ 2 3/ 22 22

11 1 2

2

d xa t t t tdt

− −⎛ ⎞= = + + − +⎜ ⎟⎝ ⎠

2

3

1 tx x

= −

5.2m 8

4m θ

v cosθ

v sinθ

Time of crosing 2

vsinθ=

Time in which truck just able to catch the man =

4

8 v cosθ−

For safe crosing 2

v sin θ = 4

8 v cosθ−or 16 2 vcos 4vsinθ θ− =

16or v

cos 2sinθ θ=

+For v minimum cos 2sinθ θ+ is maximum

( )dso, cos 2sin 0

dθ θ

θ+ =

sin 2cos 0

tan 2

1 2cos ,sin

5 5

θ θθ

θ θ

⇒ − + =⇒ =

⇒ = =

Now min

16 5 16v 3.57 m / s

5 5= = =

6. |Average velocity| = displacement

time

= 2

time 1

AB= = 2 m/s.

7. Acceleration = dv

v 0 kxdt

= = +i

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dxx v

dt

•⎧ ⎫= =⎨ ⎬⎩ ⎭∵ ( )0v a kv k v kx⇒ = = = +

i

Further, dv

a kvdt

= = dv

kvdt

⇒ =

dvkdt

v⇒ =

1

0

v t

v 0

dvk dt

v⇒ =∫ ∫

1e

0

v1t log

k v

⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠Since, v=v

0+kx. Hence slope of velocity

displacement curve is dv

kdx

=

8.l/2 l/2

u a11 u a22

21 1

l 1u t a t

2 2= + ...(1) and ( ) 2

1 2

l 1u t a t

2 2− = − + −

22 2

l 1u t a t

2 2⇒ = + ....(2)

subtracting (1) and (2) ,

we get2 1

1 2

u ut 2

a a

⎛ ⎞−= ⎜ ⎟−⎝ ⎠

....(3)

Substituting (3) in (1) or(2) and rearranging,we get.

( )( )

( )2 11 2 2 12

1 2

4 u ul a u a u

a a

−= −

− ...(4)

Since the particle P & Q reach the other ends of

A and B with equal velocities say v

For particle P 2 21 2v u 2a l− = ... (5)

For particle Q 2 22 2v u 2a l− = ...(6)

Substracting and then substituting value of l and

rearranging, we get

( )( ) ( )2 1 1 2 1 2 2 1u u a a 8 a u a u+ − = −

9.dvadt

= , If velocity changes, definiety there will

be acceleration. If speed changes, then velocityalso changes, so definietely there will beacceleration.Acceleration may be due to change in thedirection of velocity only and not magnitude.If body has acceleration, its speed may changesif acceleration is due to change in magnitude ofvelocity.

10. Average velocity

( )5 5

2

0 05 5

0 0

4vdt t t dtv

dt dt

−= =

∫ ∫

∫ ∫

�

32

0

1252 503 25 535 5 3 5 3

tt⎡ ⎤−⎢ ⎥ −

⎣ ⎦= = = =×

For average speed, let us put v= 0, which gives

0t = amd 4t s=∴average speed =

4 5 4 52

0 4 0 45

0

(4 )

5

vdt vdt t t dt vdt

dt

+ − +

=∫ ∫ ∫ ∫

∫4 53 3

2 2

0 4

2 23 3

5

t tt t⎡ ⎤ ⎡ ⎤− + −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦=

4 53 32 2

0 4 1

2 23 3 13

5 5

t tt t

ms−

⎡ ⎤ ⎡ ⎤− + −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦= =

For acceleration :

( )24 4 2dv da t t tdt dt

= = − = −

At 20, 4t a ms−= =

11.

0

0

0

0

.x

v

dvv v or dv dxdx

α α= − ⇒ = −∫ ∫

0

00 0 0

0

;

( )

αα

α α

= ⇒ =

= − = −∫ ∫v t

v

vv x x

dv dvv or dtdt v

( )0

0

0

1

α

α

−= = = ∞

⇒ = =

tv v e or v for tvv whente

.

12. The particle’s velocity is getting zero at t = 3 s,

where it changes its direction of motion.

For 0 < t < 3 s, V is negative, a is positive, so

particle is slowing down.

For t < 3, both V and a are positive, so the particle

is speeding up.

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13. Since, the body is at rest at x = 0 and x = 1. Hence,

a cannot be positive for all time in the interval 0 ≤ t ≤ 1

Therefore, first the particle is accelerated and then

retarded. Now, total time it = 1 s (given)

S = Area under v – t graph

\Height or vmax

= 2st = 2 m/s is also fixed.

max

1Area or =

2S t v⎡ ⎤× ×⎢ ⎥⎣ ⎦

2

31

O 1

V

V =

2m

/sm

ax

t

If height and base is fixed, area is also fixedIn case 2 : Acceleration = Retardation = 4 m/s2

In case 1 : Acceleration > 4 m/s2 while Retardation< 4 m/s2.Hence, a ³ 4 at some point or points in its path.

14. At time t, Xt and X

p are coordinates of train and

passenger respectively.

2t 1

1X a t

2= and p PX v t S= −

If passenger catches the train,X

t = X

p

or 2

1 P

1a t v t S

2= − or

2P P 1

1

v v 2a St

a

− −=

( ) ( )( )210 10 2 1 25.5

3seconds1

− −= =

15. The critical distacne ‘Sc’ for which passenger will

take the ten seconds time to catch the train is given

by 2P

c1

vS

2a=

The time is 10 seconds, if 2P 1v 2a S 0− =

( )( )

22P

ct

10vS 50m

2a 2 1= = =

16. For critical distance, passenger catches the train in

time, P

t

vt

a= So, required velocity of train = ta .t

Pt P

t

va V / 2 10m / sec

a

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

17. Let they meet after time t , then the distancetravelled by both in time t should be same

218 4 4

2s t t t s= = ⇒ =

18. 8 8 4 32s t m= = × =

19. 2 , 2 ,dS dvv t a So i bdt dt

β γ γ= = + = = →

( ) ( ) ( )3

23

2

3 2 9 45 , .

1

vdtv So ii a

dt

β γβ γ

− + −= = = + →

∫

∫Velocity at

, ,1 : 2 1 2 So iii dt s v β γ β γ →= = + × = +Initial displacement i.e.,

0, , ,t S a So iv c= = →

20.1 2

1 2 1 2 1 2 2 2

1 1, , ,

2 2t t t v v v S a t S a t= − = = = =

1 1 1 2 2 2 2 1 1,v a t v a t v v a t= = ⇒ + =

12 2 1 1 1 2 2

1 2

v a ta t v a t a t ta a

+⇒ + = = ⇒ =

−

( )( )

1 22 1 2

1 2 2 1 1

1 1t a aa t at

a t t a v a t−

= = − ⇒ = −+

2 22 1 2 1 2 1

11

a v a t a v a a t v a a a tv a ta

+⇒ = ⇒ + = +

+

( ) 11 2 8v a a t ms−⇒ = =

21. 1 190 25pV kmh ms− −= =1 172 20cV kmh ms− −= =

In 10 s culprit reaches point B from A. Distancecovered by culprit, 20 10 200S vt m= = × = .At time t=10s, the police jeep is 200 behind theculprit. Relative velocity between jeep culprit is

25-20 = 15 ms− .

Time 200

405

S sV

= = = ( Relative velocity is

considered )In 40 s, the police jeep will move from A to adistance S. Where

25 40 1000 1.0 .S vt m km away= = × = =The jeep will catch up with the bike 1 km farfrom the turning.

22. 2 3 26 3 4 12 3 3x t t t v t t= − − − = − −

V= max when 0 2secdv tdt

= ⇒ =

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( ) 12 9v ms−∴ = .23. According to the third equation of motion

2 2 2v u as− = Given 3 ,v v u v and a g= = =

or ( )2

2 2 43 2

vv v gs or sg

− = =

24. Let the initial relative velocity, rerlativeacceleration and relative displacement of the with

respect to the floor of the lift be ,r ru a and rs ,

then ( ) 21/ 2r r rs u t a t= +

and 10 10 0r c lu u u= − = − =

( ) 29.8 0 9.8r c la a a ms−= − = − − = −

2.45r c ls s s m= − = −

( ) ( )( ) 22.45 0 1/ 2 9.8t t− = + −

or 2 1/ 2 1 2t or t s= =25. Suppose the body be projected vertically

upwards from A with a speed 0u .

Using equation 21,

2s ut at⎛ ⎞= + ⎜ ⎟

⎝ ⎠ we get

For first case:2

0 1 1

1

2h u t gt⎛ ⎞− = − ⎜ ⎟

⎝ ⎠( i )

For second case: 2

0 2 2

1

2h u t gt⎛ ⎞− = − − ⎜ ⎟

⎝ ⎠( ii )

( i )- ( ii ) ( ) ( )2 20 2 1 2 1

10

2u t t g t t⎛ ⎞⇒ = + + −⎜ ⎟

⎝ ⎠

( )0 1 2

1

2u g t t⎛ ⎞⇒ = −⎜ ⎟

⎝ ⎠( iii )

Putting the value of 0u in ( ii ), we get

( ) 21 2 2 2

1 1

2 2h g t t t gt⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1 2

1

2h gt t⇒ = ( iv)

For third case: 0, ?u t= =

210

2h t gt⎛ ⎞− = × − ⎜ ⎟

⎝ ⎠ or

21

2h gt⎛ ⎞= ⎜ ⎟

⎝ ⎠( v)

Combining Eq. ( iv ) and ( v), we get

21 2

1 1

2 2gt gt t= or 1 2t t t=

26. Suppose h be the height of each storey, then

2 21 125 0 10 10 5 5

2 2h t or h m= + × × = × × =

In first second, let the stone passes through nstorey.

So ( )215 10 1 1

2n or n× = × × =

27. ( )2 21 1;0

2 2r r rS u t a t ut g a t= + = − +

28. i) For uniformly accelerated/decelerated motionv2= u2 ± 2ghi.e., v – h graph will be a parabola Becauseequation is quadratiC).ii) Initially velocity is downwards (–ve) and thenafter collision it reverses its direction with lessermagnitude. I.e., velocity is upwards (+ve). GraphA) satisfies both these conditions.Therefore, correct answer is A).Note that time t = 0 corresponds to the point onthe graph where h = d

d3 1

2

2

h

v

at t = 0, h = d

Collision takesplace here

1 2: increasesdownwards

At 2 velocity changesits direction

3V decreases upwards

→

→

→ 2 29. Distance traveled in tth second is,

st = u + at – 1

2a Given : u = 0

∴ 1

n

n

ss +

= ( )

1–

21

1 –2

an a

a n a+ =

2 – 1

2 1

nn +

Hence, the correct option is C).

30. As going up, speed of the particle is decreasing

and hence time taken in crossing the windows

(if 1 2 3S S S= = ) will be 1 2 3t t t .< < Since,

v u at= +� � �

u t∆ ∝ As acceleration is same)

So, 1 2 3v v v∆ < ∆ < ∆ As for equal windows

1 2 3t t t< < ) For unequal windows.

1 2 3t t t= = if 3 2 1S S S< <31. Let they meet at hieght h after time l .

21100 bullet

2h t gt for first= − →

( ) ( )21100 5 5

2t g t= − − − → for second bullet

12.5t s⇒ = − ( after solving) So ( s) is correct.Time of flight of first bullet

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2 2 10020

100

uT sg

×= = =

Second bullet will reach after 5 s of reaching first. So ( B) is correct.

11 100 10 20 100v ms−= − × = −

12 100 10 15 50v ms−= − × = −

ratio: 1

2

2 :1vv

= So ( C) is correct.

maximum height attained

( )22 100500

2 2 10

uH mg

= = =×

.

Hence ( d ) is incorrect32. Ball A will return to the top of tower after

2 2 102

10

uT sg

×= = =

With speed of 110ms− downward.And this time, B is also projected downwards with

110ms− .So both reach ground simultaneously. Also theywill hit the ground with the same speed.

33. ( )2x t 3= − 2x t 6t 9⇒ = − +

dxv 2t 6

dt⇒ = = −

At the point of reversal, v=0,so 2t 6 0− = t 3s⇒ =

t 0 1 2 3 4 5 6x 9 4 1 0 1 4 9v -6 -4 -2 0 2 4 6a 2 2 2 2 2 2 2

Nature of

Motion

Decele rating

Decele rating

Decele rating

Accele rating

Accele rating

Accele rating

Accele rating

Displacement of the particle from t =0 to t =6s is

zero Distance travelled by the particle from t =0 to

t=6 is 9+9=18m Average speed of the particle is

av

TotalDistanceTravelledv

TotalTimeTaken=

1av

18v 3ms

6−⇒ = =

Average Velocity of the particle is

av

Displacementv 0

Time= =�

Acceleration of the particle over the entire duration

of motion is 2ms-2.

Displacement of the particle from t =0 to t =6s is

zero Distance travelled by the particle from t =0 to

t=6 is 9+9=18m

Average speed of the particle is

av

TotalDistanceTravelledv

TotalTimeTaken=

1av

18v 3ms

6−⇒ = =

Average Velocity of the particle is

av

Displacementv 0

Time= =�

Acceleration of the particle over the entire durationof motion is 2ms-2.

34. For a round trip, displacement is zero; hence

0.avv =�

Also

1 21,

2av

v vv when v+=� ��

is initial, 2v�

is final.

Hence . ,i a b→Average speed

( avv ) = ( )2

0 0

0

2 / 2tan

2 / 2

v g vTotal dis ceTime of flight v g

= =

0 .,ascent descentvT T Henace iii d iv dg

= = → →

35. Taking upward direction as positive, let us workin the frame of lift. Acceleration of ball relative to

lift = ( g+ A) downwards, so ( )reala g a= − + ,

initial velocity; relu v= , final velocity relv v= −as the ball will reach the man with same speedw.r.t lift.

Apply ( ) 2rel rel relv u a t v v g a t t s= + ⇒ − = + − − ⇒ =

36. ( )2 12

as u n= + −

1 2100 , 10 5u ms a ms and s m− −= = − =

( )5 10 5 2 1 10n gives n s= − − =

Body when thrown up with velocity 1200 ms−

will take 20 s to reach the highest point. Distancetravelled in 20th second is

( )200 5 20 2 1 5 m− × − =In the last second of upward journey , the bodieswill travel same distance.

37. Maximum height will be attained at 110 s.Because after 110 s, velocity becomes negative

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and rocket will start coming down. Area from 0

to 110 s is1

110 1000 55,000 552

m km× × = =

38. Acceleration between 8 to 10 s( or at t= 9 s) .

22 1

2 1

5 155 /

10 8

v va m st t

− −= = = −

− −39. Particle changes direction of motion at .t T=

Acceleration remains constant, because thevelocity-time graph is a straight line. Displacementis zero, because net area is zero. Initial and finalspeeds are equal.

40. Maximum value of position coordinate = initialcoordinate+area under graph up to 24t s= ( As up tot= 24 s, the displacement of the particle will be positive )

6

2

0–2 10 18 24

30 40t(s)

V

–6

Maximum value of position coordinate

( ) ( ) ( )2 6 1 616 2 10 18 10 24 18

2 2

⎡ + × ⎤⎛ ⎞− + ×× + × − + × −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

41.50

10

x tt −=

2 2 25010 50 10 50

10

x tt t x t t t−= ⇒ = − ⇒ +

Instantaneous velocity 20 50dxV tdt

= = +

average velocity

( ) ( )3

30

0

20 50avg

Vdt t dtV

dt dt

+= =∫ ∫∫ ∫

( )

( )

323

00

3

0

20 502

avg

t tV

t

⎛ ⎞+⎜ ⎟

⎝ ⎠=

( )20 950 3 90 1502

3 3avgV

×+ +

= =

124080 80

3avg avgV V ms−= = ⇒ =

42. Let ‘t’ be the time at which A overtakes B.For A: x-50=V

At; ForB: x-100=V

Bt

x=50+VAt ; x=100+V

Bt

50+VAt=100+V

Bt

⇒ 50+10t=100+5t ⇒ 5t=50

t = 10 secx=50+V

At=50+10x10=150⇒ 150m

43. Let t1 be the time of freefall.

t2 be the time of travel of sound

1

2 0 67.6 676 26

9.84.9 49 7

htg

×= = = =

t1 = 3.71 sec

Time of travel of sound t2 = 3.91-3.71=0.2 sec

Velocity of sound ( ) ( )( )

distance 67.6338

time 0.2

dV

t= = =

1338V ms−=44. Velocity after 10 sec V=gt

1=10x10=100ms–1

Distance travelled free fall

21 1

1 110 100

2 2d gt= = × ×

d1 = 500m

Distance travelled with retardaton

( )2 2

12

25 10000 9975

2 2 2.5 5

V Vda

− −= = =

−

2 1995d m=Height from which he bails out is

1 2 500 1995 2495h d d m= + = + =2495h m=

A. Motion along Horizontal AxisSINGLE QUESTION TYPE

1. The decelaration experienced by a movingmotor boat, after its engine is cut-off is given

by 3dv kvdt

= − , where k is constant.If 0v isthe magnitude of the velocity at cut-off, themagnitude of the velocity at a time t afterthe cut-off is

A) 0 / 2v B) v C) 0ktv e− D)

0

202 1+

vv kt

2. A jet plane starts from rest at S = 0 and issubjected to the acceleration shown.Determine the speed of the plane when it hastravelled 60 m.A) 46.47 /m s

a(m/s)2

150 S(m)

22.5

B) 36.47 /m s

C) 26.47 /m s

D) 16.47 /m s

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Jr|11th|I PUC IIT-JEE ADVANCED|PHYSICS:VOL-I3. Velocity versus displacement graph of a

particle moving in a straight line is shown infigure. Corresponding acceleration versusvelocity graph will be :

v(m/s)

10

10 s(m)

A)

a(m/s )2

10

10 v(m/s)

B)

a(m/s )2

10

10 v(m/s)

C)

a(m/s )2

10

10 v(m/s)

D)

a(m/s )2

10

10 v(m/s)

4. The motion of a body falling from rest in aresisting medium is described by the equationd v

= a-b vd t

where a and b are constants.

The velocity at any time t is given by

A) ( )-b tav= 1 -e

bB) ( )-btb

v= ea

C) ( )-b tav= 1+e

bD)

btbv= e

a5. Two stones are thrown up simultaneously with

initial speeds of u1and u2 (u2>u1). They hit the

ground after 6 s and 10 s respectively. Whichgraph in fig.correctly represents the timevariation of 2 1x (x x )∆ = − , the relativeposition of the second stone with respect tothe first upto t=10 s? Assume that the stonesdo not rebound after hitting the ground.

A)

∆x

B

02 4 6 8 10

t

B)

∆x

B

02 4 6 8 10

t

A

C)

∆x

B0

2 4 6 8 10t

A

D)

∆x

B0

2 4 6 8 10t

A

6. A cone falling with a speed 0v strikes andpenetrates the block of a packing material.The accelertation of the cone after impact is

2a g cx= − . Where c is a positive constantand x is the penetration distance. If maximumpenetration depth is mx then c equals

A) 20

2

2 m

m

gx vx+

B) 20

2

2 m

m

gx vx−

C) 20

3

6 3

2m

m

gx vx−

D) 20

3

6 3

2m

m

gx vx+

MULTIPLE ANSWER QUESTIONS7. If the velocity of the particle is given by

v x= and intially particle was at x=4m, thenwhich of the following are correct.A) at t=2 sec, the position of the particle is x= 9 mB) Particle acceleration at t= 2 sec. is 1 m/s2.C) Particle acceleration is ½ m/s2 through out themotionD) Particle will never go in negative direction fromits starting position.

8. Starting from rest a particle is first accelerated fortime t1 with constant acceleration a1 and then stopsin time t2 with constant retardation a2. Let v1 bethe average velocity in this case and s1 the totaldisplacment. In the second case, it is acceleratedfor the same time t1 with constant acceleration 2a1and comes to rest with constant retardation a2 intime t3. If v2 is the average velocity in this caseand s2 the total displacement. ThenA) v2 = 2v1 B) 2v1 < v2 < 4v1C) s2 = 2s1 D) 2s1 < s2 < 4s1

PASSAGE TYPE QUESTIONSComprehension - 1Two particles A and B start from rest at theorigin x=0 and move along a straight line suchthat aA=(6t-3)ms-2 and aB=(12t2-8)ms-2, wheret is in seconds. Based on the above facts.Answer the following questions.

9. Total distance travelled by A at t=4 s isA) 40m B) 41m C) 42m D) 43m

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10. Total distance travelled by B at t=4 s isA) 192m B) 184m C) 196m D) 200m

11.. Total distance between them at t = 4 s isA) 144m B) 148m C) 152m D) 156mComprehension - 2A balloon is start rising with constantacceleration 2m/s2 from ground at t=0s. Astone is dropped at t=5s. s-t graph for the givensituation is shown in figure. Answer the following.s

5s t1 t2

t

12. The maximum hight reached by the stone isA) 30m B) 40m C) 45m D) 28m

13. t1 isA) 4s B) 6s C) 2s D) 1s

14. t2 isA) 4s B) 4.45s C) 3.45s D) 9.45s

MATRIX MATCHING QUESTIONS15. Study the following.

List - IA) A body covers first half of distance with a speedv1 and second half of distance with a speed v2B) A body covers first half of a time with a speedv1 and second half of time with a speed v2C) A body is projected vertically up from groundwith a speed gh .Considering its total motionD) A body freely released from a height hList - II

i) Average speed is 2

gh

ii) Average speed is 1 2

2

v v+

iii) Average speed is 1 2

1 2

2v vv v+

iv) Average speed is 2

gh

16. For a particle moving along X - axis ifacceleration Constant) is acting along -ve X-axis, then match the entires of Column I withentires of Column II.Column -IA) Initial velocity > 0 B) Initial velocity < 0C) x > 0 D) x < 0

Column -IIi) Particle may move in +ve X - direction withincreasing speed.ii) Particle may move in +ve X-direction withdecreasing speed.iii) Particle may move in -ve X - direction withincreasing speed.iv) Particle may move in -ve X - direction withdecreasing speed.

INTEGER TYPE QUESTIONS17. A train starts from station A with uniform

acceleration a1 for some distance and thengoes with uniform retardation a2 for somemore distance to come to rest at station BThe distance between stations A and B is 4km and the train takes 1/15 h to completethis journey. Acceleration are in km/ 2min

unit. If 1 2

1 1 xa a

+ = , find the value of x.

18. A cat, on seeing a rat a distance 5 ,d m=

starts with velocity 15u ms−= and moveswith acceleration 22.5 msα −= in order tocatch it, while the rat with acceleration βstarts from rest. For what value of β will thecat overtake the rat ? (in ms–2)

19. On a two-lane road, car A is travelling with aspeed of 36kmh–1. Two cars B and C approachcar A In opposite directions with a speed of54kmh–1 each. At a certain instant, when thedistance AB is equal to AC, both being 1 km,B decides to overtakes A before C does. Whatminimum acceleration of car B (in m/s2) isrequired to avoid an accident ?

B A C1000 m 1000 m

v = 15msB

–1 v = 10msA

–1 v = 15msC

–1

20. The accelerator of a train can produce a uniformacceleration of 0.25 ms-2 and its brake canproduce a retardation of 0.5 ms-2 . The shortesttime in which the train can travel between twostations 8 km apart is x minutes and 10 s, if itstops at both stations. The value of x is.

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21. A body starts from rest with uniformacceleration. Its velocity after 2n second isv0. the displacement of the body in last n

second is 03v nβ . Determine the value of β ?

B. Motion Along Vetical Axis :SINGLE ANSWER QUESTIONS

22. A ball is thrown from the top of a tower invertically upward direction. Velocity at apointh metre below the point of projection is twiceof the velocity at a point h metre above thepoint of projection. find the maximum heightreached by the ball above the top of tower.A) 2 h B) 3 h C) (5/3) h D) (4/3) h

23. A parachutist drops first freely from anaeroplane for 10 s and then his parachuteopens out. Now he descends with a netretardation of 2.5ms–2. If he bails out of theplane at a height of 2495 m and g=10ms–2,his velocity on reaching the ground will beA) 15 ms− B) 110 ms− C) 115 ms− D) 120 ms−

MORE THAN ONE QUESTION TYPE24. A particle is thrown vertically in upward

direction and passes three equally spacedwindows of equal heights thenA) average speed of the particle while passing thewindows satisfies the relation

1 2 3av av avu u u> >

B) the time taken by the particle to cross thewindows satisfies the relation t

1 < t

2 < t

3

C) the magnitude of the acceleration of the particlewhile crossing the windows satisfies the relation a

1= a

2 ≠ a

3

D) the change in the speed of the particle whilecrossing the windows would satisfy the relation

1 2 3u u u∆ < ∆ < ∆ .

COMPREHENSION QUESTIONAn elevator car whose floor to ceiling distancein equal to 2.7 m starts ascending with constantacceleration 1.2m/s2, 2 sec. after the starts abolt begins falling from the ceiling of the car.Answer the following questions. (g=9.8 m/s2)

25. The bolt’s free fall timeA) 0.3 s B) 0.5 s C) 0.7 s D) 0.9 s

26. The velocity of bolt at instant it loses contact isA) 1.2 m/s B) 2.4 m/s C) 4 m/s D) 10 m/s

27. Distance moved by elevator car w.r.t. groundframe during the free fall time of the bolt.A) 1.44 m B) 1.63 m C) 1.68 m D) 1.97 m

28. Distance covered by the bolt during the freefall time w.r.t. ground frame.A) 0.7 m B) 0.9 m C) 1.1 m D) 1.3m

29. The displacement by the bolt during its freefall time w.r.t. ground frameA) 0.3 m B) 0.7 m C) 0.9 m D) 1 m

INTEGER TYPE QUESTIONS30. A stone is dropped from a height h.

Simultaneously another stone is thrown upfrom the ground with such a velocity that itcan reach a height of 4h. The time when two

stones cross each other is hkg

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠ where k =

31. A particle moves along x-axis satisfying theequation ( )( )1 2x t t t⎡ ⎤= − −⎣ ⎦ (t is in secondsand ‘x’ is in meters). Find the magnitude ofinitial velocity of the particle in m/s.

32. The position vector of a particle varies withtime as ( )0 1r r t tα= −

��where 0r

� is a constant

vector and α is a positive constant. Thedistance travelled by particle in a time intervalin which particle returns to its initial position

is 016

Krα

. Determine the value of K?

C. Graphs:SINGLE ANSWER TYPE

33. The velocity-time graph of abody is given infigure. The maximum accceleration in ms–1 is

A) 4

60

20

020 30 40 70

t(s)

v(ms )–1

B) 3

C) 2

D) 1

34. The velocity-time graph of a body is shown infigure. The ratio of magnitude of averageacceleration during the intervals OA and AB is

A) 1

30° E 60°A BO t(s)

CD40

v(ms )–1

B) 1/2

C) 1/3

D) 3

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35. The displacement-time graph of a movingparticle with constant acceleration is shown inthe figure. The velocity time graph is given by

x(m)

t(s)0 1 2

A)

v

1 20

t B)

v

120

t

C)

v

1 20t D)

v

1 20t

MULTIPLE ANSWER QUESTIONS36. The velocity-time plot a particle moving on

a straight line is shown in figure.

10

10

0

–10

–20

v(ms )–1

20 30t(s)

A) The particle has a constant acceletrationB) The particle has never tuned aroundC) The particle has zero displacementD) The average speed in the interval 0 to 10 s is thesame as the average speed in the interval 10 s to 20 s

37. The displacement of a particle as a functionof time is shown in figure. It indicates

S

O1 2 4 5

t

A) The particle starts with a certain velocity, butthe motion is retarded and finally the particle stopsB) The velocity of the particle decreasesC) The acceleration of the particle is in oppositedirection to the velocity

D) The particle starts with a constant velocity,the motion is accelerated and finally the particlemoves with another constant velocity.

PASSAGE TYPE QUESTIONPassage-1:

Study the following graphs:

i)

x

t

ii)

x

t

iii) x

t

iv) x

t

38. The particle is moving with constant speedA) In graphs (i) and (iii) B) In graphs (i) and (iv)C) In graphs (i) and (ii) D) In graphs (i)

39. The particle has negative accelerationA) In graph (i) B) In graph (ii)C) In graph (iii) D) In graph (iv)

MATRIX MATCHING QUESTION40. The displacement versus time is given

figure. Sections OA and BC are parabolic.CD is parallel to the time axis.

C D

BA

O

S

t

Column-IA. OA B. AB C. BC D. CDColumn-IIi. Velocity increases with time linearlyii. Velocity decreases with timeiii. Velocity is independent of timeiv. Velocity is zero

41. Study the following v t− graphs in Column Icarefully and match appropriately with thestatements given in Column II. Assume thatmotion takes place from time 0 to T.Column-I

A) Tt

v

O

–v0

B) T t

v

O

v0

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C) T

v

O

–v0

T/2

v0

D) T t

v

O

v0

i

Column-IIi. Net displacement is positive, but not zeroii. Net displacement is negative, but not zeroiii.Particle returns to its initial position againiv. Acceleration is positive.

NUMERICAL GRADING42. A mouse can crawl with a velocity 2m/s on a

stationary truck of rectangular shape. Thetop view of the truck is shown in the dig. Thetruck is moving with constant speed 5 m/salong a straight road. The points P, Q, B andC are the corners of the truck and A ismidpoint of PQ. Now the mouse startsmoving from point A and following the cyclicpath A B C A→ → → along the triangularmark shown. What is the displacement(inmeters) of mouse with respect to groundwhen it returns to point A? [Assume that themouse moves with 2m/s w.r.t. truckthroughout the motion]

A

BC

P Q

5 m/s4 m

6 m

43. A particle initially at rest moves along x-axis.Its acceeleration varies with time asa=(6t+5)m/s2. The distance covered in first 2sec is (in meters)

44. An object is dropped from the top of a tower.It travels a distance ‘x’ in the first secodnand 7x in the last second. If g=10m/s2, thenthe height of tower is (in meters)

45. Two paper screens A and B are separated bya distance of 100 m. A bullet pierces A andB. The hole in B is 10 cm below the hole in A.If the bullet is travelling horizontally at thetime of hitting A, then g=9.8m/s2 the velocityof the bullet at A is (in m/s)

46. A stone is dropped from the top of a tower20 m high. Simultaneously another stone isthrown so that it can reach the top of thetower. What is the sum of speeds of thestones while they pass one another? (In m/s)

1) D 2) A 3) A 4) A 5) A 6) D7) A,C,D 8) A,D 9) B 10) D 11) C 12) A13) B 14) D15) A-iii, B- ii, C-iv, D-i16) A-ii, B-iii, C-ii, D- ii,iii17) 2 18) 5 19) 1 20) 5 21) 4 22) C23) A 24) ABD 25) C 26) B 27) D28) D 29) B 30) 8 31) 2 32) 8 33) A34) C 35) A 36) AD37) ABC 38) B39) C40) A - i, B - iii, C - ii, D - iv41) A - ii,iv, B - i,iv C - iii, D - i42) 0040 43) 0018 44) 008045) 0700 46) 0020

1. Here 3dv kv

dt= −

or 0

3 3 0

v t

v

dv dvkdt or kdtv v

= − = −∫ ∫

or 0

2 2 20

1 1 2

2 2 2

v

V

kt or ktv v v

⎡ ⎤− = − − + = −⎢ ⎥⎣ ⎦ , or

22 0 0

2 20 0

1 2 2 1

v vv or vv kt v kt

= =+ +

2.dv

a vds

= s v

2 2

0 ua.ds v.dv v u⇒ = = −∫ ∫

⇒ Area under a s∼ curve 2 2v u= −

⇒ ( ) 21 1150 22.5 90 13.5 v 0

2 2⎛ ⎞× × − × = −⎜ ⎟⎝ ⎠

2v 75 22.5 45 13.5⇒ = × − ×v 75 22.5 45 13.5⇒ = × − × 46.47⇒

4.dv

a bvdt

= −v t

0 0

dvdt

a bv=

−∫ ∫

( ) v

0

2ln a bv t

b

−⎛ ⎞ − =⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠

a bvln bt

a

−⇒ = −

bta bv ae−⇒ − = ( )btav 1 e

b−⇒ = −

5.2

1 1

1

2x u t gt= − ;

22 2

1

2x u t gt= −

It is valid up t=6 sec only

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6.2dvv g cx

dx= −

0

02

0 0

x x

v

vdv g dx c x dx⇒ = −∫ ∫ ∫2 30

2 3

v cxgx⇒ − = −

2 230 0

2 3

33

2 2

v vx c ggx cc x x

⇒ = + ⇒ = +

7. v x=

1/22

dx dxx dt x t cdt x

= ⇒ = ⇒ = +∫ ∫0, 4, 4But at t x c= = ⇒ =

( ) ( )2 24 6 36

94 4 4

tx x

+⇒ = ⇒ = = =

21½ / .

2

dva v x m sdx x

= = × =

8. Valid relations are : v2=2v

1; 2s

1<s

2<4s

1

9,10,11 For A

A Adv a dt=0 0

(6 3)AV t

Adv t dt⇒ = −∫ ∫23 3AV t t⇒ = −

For B

B Bdv a dt= 0

2

0 0

(12 )V t

Bdv t B dt⇒ = −∫ ∫34 8BV t t⇒ = −

Let us now calculate the times when A and Bare at rest. The particle A is at rest (V

A=0),

when 3t2-3t=0 ⇒ t = 0 s and t = 1 sThe particle B is at rest (V

B=0), when

4t3 - 8t = 0 ⇒ t = 0 s and t = 2 sThe position of particles A and B can be detemined

using ,dxv sodt

= dxA= v

A dt

2

0 0

(3 3 )AX t

Adx t t dt⇒ = −∫ ∫3 23

2AX t t⇒ = − Similarly dXB = v

Bdt

3

0 0

(4 8 )BX t

Bdx t t dt⇒ = −∫ ∫ ⇒ XB = t4 - 4t2

The positions of particle A at t = 1 s and 4s are

3 21

31 (1 ) 0.5

2A t sX m= = − = ; 3 24

34 (4 ) 40

2A t sX m= = − =

Particle A has travelled a total distance given byd

A=2(0.5)+40=41m

The positions of particle B at 2t = s and 4s

are ( ) ( )4 2

22 4 2 4B tX m

== − = −

( ) ( )4 2

4 4 4 4 192B tX m= = − =

Particle B has travelled a total distance given byd

B = 2(4)+192=200m

At t = 4 s the distance between A and B is

192 40 152ABX m∆ = − =

12. 21; ; 0

2s ut at a g v= + = − =

13. At 1; 0t t v= = 14. At 2; 0t t s= =15. Equation of motion 16. Equation of motion

17. 1 2 1 1 2 24 min,t t v a t a t+ = = =1

4 4 2 22

S v v v= × ⇒ = ⇒ =

v

1a 2a

1t 2tt

1 2

1 2 1 2 1 2

1 1 1 1 1 14 2 2t t v

a a a a a a⎡ ⎤⎡ ⎤

+ = + ⇒ = + ⇒ + =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

18. For rat 21

2S tβ= ( i )

For cat 210

2S d ut tα= = + ( ii )

Putting the value of S from Eq. ( i ) in Eq ( ii ),

( ) 2 2 2 0a b t ut d− + − =

( )22 4 8 (

2

u u dt

β αβ α

± − −=

−

For t to be real, ( )2

2

ud

β α≥ −2

2

ud

β α∴ = +

Substituting a, d and u we get2

252.5 2.5 2.5 5

2 5msβ −= + = + =

×19. The situation can be roughly shown in the figure.

Let C take time t to overtake A.

B A C1000 m 1000 m

v = 15msB

–1 v = 10msA

–1 v = 15msC

–1

( ) 11000 , 10 15 25rel reld m v ms−= = + =

Here 1000

4025

rel

rel

dt sv

= = =

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Jr|11th|I PUC IIT-JEE ADVANCED|PHYSICS:VOL-ILet acceleration of B be a for overtaking.

11000 ; 15 10 5rel reld m v ms−= = − =

40reld a and t s= =

Using 21

2rel rel reld u t a t= +

( )2 21100 5 40 40 1

2a a ms−= × + ⇒ =

20.( ) 1/2

min

2 +⎡ ⎤= ⎢ ⎥

⎣ ⎦

lt

α βαβ

1

2

max

2 lv αβα β

⎛ ⎞= ⎜ ⎟+⎝ ⎠

( ) 3

min

0.25 0.5 8 10 2310 5min10

0.25 0.5

+ × ×= = =

×t s s .

21. 0v 2na= 0va

2n= ( )2 21 1

s 2n an2 2

= −

23an

2=

20 0v 3v n3n

2 2n 4= =

22.2

2 1; 22

uH given v vg

= =

i) A to B: 2 21 2v u gh= −

ii) A to C ( )2 22 2v u g h= − −

iii) solving (i), (ii),(iii) we get the value of 2u as 10g/h/

3 and then we get the value of H by using 2

2

uHg

=

h

h

Hv1

B

u

A

C

v2

23. The velocity v acquired by the parachutist after 10 s.10 10 10 100v u gt ms−= + = + × =

Then, 2 2

1

1 10 10 10 500

2 2s ut gt m= + = + × × =

The distance travelled by the parachutist underretadation is

2 2495 500 1995s m= − =Let gv be the velocity on reaching the ground.

Then 2 222gv v as− =

or ( ) ( )22 1100 2 2.5 1995 5g gv or v ms−− = × − × =

24. As the particle is going up, it is slowing down,i.e., speed is decreasing and hence we can saythat time taken by the particle to cover equaldistances is increasing as the particle is going up.Hence, t

1 < t

2 < t

3.

As av

Distanceu ,

time= we have av

1u

time∝

Acceleration throughout the motion remains samefrom equation,

v u at,= +��

u t∆ ∝ . So, 1 2 3u u u∆ <∆ < ∆ .

25. ( )2 21 12.7 0 9.8 1.2 0.7sec

2 2rel rel rely u t a t t t= + − = − + =

26. v = u + at = 0+(1.2)(2) = 2.4 m/s

27. ( )( ) ( ) ( )221 12.4 0.7 1.2 0.7

2 2y ut at= + = +

28. Distance( )22 2.42

| | 0.7 1.32 9.8

us y mg

∆= + = + =

29. b be eS S S= +� � �

2.7 1.97bS = − + 0.7bS m≈ .

30.2 2

1 2

1 1;

2 2S gt S ut gt= = −

2

1 2 ; 4 82

uS S h h u ghg

+ = = ⇒ =

88

hut h ght h tg

⇒ = ⇒ = ⇒ =

31. ( )2 3 23 2 3 2x t t t t t t= − + = − +23 6 2;v t t= − + 0 2v⇒ =

32. ( )0 1r r t tα= −

0 02v r trα= −

0v∴ = at 1

2t

α=

∴ distance travelled before coming to originalposition

24 2

r rrα α

⎡ ⎤∴ = =⎢ ⎥

⎣ ⎦8k∴ =

33. Maximum acceleration will befrom 30 to 40 s,because slope in this interval maximum

22 1

2 1

60 204

40 30

v va mst t

−− −= = =

− −

34. During OA , acceleration = 0 21

tan 303

ms−=

During AB , acceleration = 0 2tan 60 3 ms−− = − .

required ratio = 1 / 3 1

33=

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35. At 0t = , slope of the x-t graph is zero; hence, velocityis zero at 0t = . As time increases, slope increases innegative direction; hence, velocity increases in negativedeirection. At point’I’, slope changes suddenly fromnegative to positive value: hence, velocity changessuddenly from negative to positive and then velocitystarts decreasing and becomes zero at’2’, option ( A)represents all these clearly.

36. Since the graph is astraight line, its slope is constant,it means acceleration of the particle is constant.Velocity of the particle changes from positive tonegative at 10t s= , so particle changes directionat this time.The particle has zero displacement up to 20 s,but not for the entire motion.The average speed in the interval of 0 to 10 s isthe same as the average speed in the internal of10 s to 20 s because distance covered in bothtime interval is same.

37. Initially at orogin, slope is not zero, so the particlehas some initial velocity but with time we see thatslope is decreasing and finally the slope necomeszero, so the particle stops finally.PASSAGE TYPE

38,39. 1) For the graphs ( i ) and ( iv ), slope isconstant , hence the velocity is constant2) For The graph ( iii ), the particle’s velocity firstdecreases and then increases in negative directio. Itmeans negative acceleration is involved in this motion.

40. In 2, , 2

dSOA S t v tdt

∝ = ∝

i. e., v t∝i.e., velocity increases with time

In , , 1dSAB S t vdt

∝ = ∝

41. 1) Area of v t− graph lies below the time axis,so displacement is negative, but slope is positive,so accceleration is also positive.2) Area of v t− graph lies above the time axis,so displacement is positive, and slope is positive,so accceleration is also positive.3) Displacement is zero, because half area isabove time axis and half below. Slope is negative,so acceleration is negative4) Area of v t− graph lies above the time axis,so displacement is positive, and alope is negative,so acceleration is also negative.

42. Time taken to reach A is AB BC CA

v+ +

=

5 6 58sec

2t + +

=

displacement is 5 x 8 = 40m

43. a=6t+5

6 5dv tdt

= +

( )0 0

6 5v tdv t dt= +∫ ∫

226

5 3 52

tv t t t⎛ ⎞

= + = +⎜ ⎟⎝ ⎠

23 5ds t tdt

= +

( )2

0 03 5

s tdx t t dt= +∫ ∫

22 2

0

3 5 208 0

3 2 2

t tS⎡ ⎤ ⎡ ⎤= + = + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

S = 18m

44. ( )211

2x g=

( )... 12

gx =

( ) ( )7 2 1 .... 22

gx n= −

From (1) & (2)

( )7 2 12 2

g g n= −

2n-1= 7n = 4 sec∴ h = 80 m

45.100tV

=

21 100

0.1 9.82 V

⎛ ⎞= × ×⎜ ⎟⎝ ⎠

700 /V m s∴ =46. Let they meet after time t, and let the speed with

the stone is thrown up be u. then2

2 20 /2

uh u gh m sg

= ⇒ = =

Now 2 21 1

20 20 1 .2 2

t gt gt t s− + = ⇒ =

Hence they pass other after 1s speed of the stoneprojected up = 20-10 x 1 = 10m/sspeed of the stone dropped = 0+10x1=10 m/s.sum of the speeds of the stones while they passone another is 10+10=20

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