Chapter 2

Motion in a Straight Line

DRAHMOST
Part-A (JEE Main)
5. A
cyclist starts from the point Pofa circular
Topic1-Distance, Displacement & ground ofradius 2km and travels along its
Uniform Motion circumference to the point S. The
displacement of acyclist is:
I Multiple Choice Questions |April 4. 2024 (II)
(a) 6km (b) v8 km (c) 4km (d) 8km
1. Which of the following curves possibly represent 6.
one-dimensional motion of aparticle ? [April 3, 2025 (D) A person travels x distance with velocity v, and then x
distance with velocity v, in the same direction. The average
(A) Phase (B) Velocity velocity of the person is v, then the relation between v, v,
and v, will be : [April 10, 2023 (1L))

Time (a) v=v, t, (b) v=

2
displacement
(c)
2_1,1 (d) V V
(C) Velocity (D) Total distance 7 car travels a distance of 'x' with speed V, and then same
A
distance 'x' with speed V, in the same direction. The average
speed of the car is: {Jan. 25. 2023 (1))
Time
Time (a) (b)
2(v, +v2) 2
2x
Choose the correct answer from the options given below: (c) (d)
2vjv2
(a) A, B and D only VË +V
(b) A, B and C only 8. A particle is moving with constant speed in a circular path.
(c) Aand Bonly (d) A, C and D only When the particle turns by an angle 90°, the ratio
2 A sportsman runs around a circular track ofradius r such of
that he traverses the path ABAB. The distance travelled
instantaneous velocity to its average velocity is t:xy2.The
value of x will be
and displacement, respectively, are [April 2, 2025 (11) |April 6. 2023 (1))
(a) 2 (b) 5
(a) 2r,37
(c) 1 (d) 7
(b) 3ru,u A 9. As showm in the figure, a particle is moving
"B
(c) Tu,3r speed m/s. Considering its motion fromwith constant
A to B, the
(d) 3ru, 2r magnitude of the average velocity is: |April 6. 2023 (11))
3. (a) Tm/s B
Aparticle moving in a straight line covers halfthe
with speed 6 m/s. The other half is covered in two distance (b) 3 m/s
time intervals with speeds 9 m/s and 15 m/s equal 120°
respectively.
The average speed of the particle during the motion is : (c) 2/3 m/s
(a) 8.8 m/s
|April 9, 2024 (I)) (d) 1.5V3 m/s
(b) 10 m/s (c) 9.2 m/s (d) 8 m/s 10,
4. A train starting from rest first accelerates An object moves with speed v,, v, and '; along a
line
speed of 80 km/h for timet, then it movesuniformly up to a
with a constant
segment AB, BCand CD respectively as shown in figure.
speed for time 3t. The average speed of the train for Where AB= BC and AD = 3AB, then average speed of the
this object will be:
duration of journey will be (in km/h): |April 6, 2024 ()| Feb. 1, 2023 (|
(a) 80 (b) 70 (c) 30 (d) 40 A B
A18

journey trom centre

Physics
19. A person starts his
(a) (b) *0' ofthe park and comes back to the same
3( yvy + vvy +V34) position following path OPQO as
shown

(yty tv) (y ty t3) in the figure. The radius of path taken by

(c) (d) the person is 200 m and he takes 3
min 58
3
11. A body is moving with constant speed, in a circle of radius sec to complete his
10m. The body completes one revolution in 4 s. At the journey. The average speed of the person is
of 3rd second, the displacement of body (in m) fromend its (taken =3.14) June ms
starting point is: 30. 2022 (
|Jan. 31, 2023 (II)1 20. Acar covers AB distance with first one -third at
(a) 30 (b) 1Sr
(c) Sn (d) 10/2
12. A vehicle travels 4 km with speed of 3km/h and
ms, second onethird at V, ms and last velocity v,
one-third
another ms'. Ifv, =3v,, V, 2v, and v, =1l ms then the
4 km with speed of5 km/h, then its average
speed is : velocity of the car is ms average
|Jan. 30, 2023 (ID)
(a) 4.25 kmh (b) 3.50 km/h
(c) 4.00 kmh (d) 3.75 km/h
A B |June 28, 2022 (11)1
13. Aperson moved fromn A to B on a circular path as
figure. If the distance travelled by him is 60 m, shown in
then the
magnitude of displacement would be: |July 25. 2022 (D)
(Given cos135° =-0.7) Topic 2-Non-uniform Motion
(a) 42 m
(b) 47m 135°
A
(c) 19m Multiple Choice Questions
(d) 40m 21. The displacement xversus time graph is shown
below.
14. A particle is moving with speed
y=bx along positive dis(m)
placement
x-axis. Calculate the speed of the particle at time 10+
that the particle is at origin at t 0). t=t (assume
|12 Apr. 2019 (D) 5

(a) b'r
4
(b) (c) b (d)
2 5 time (s)
15. All the graphs below are
motion. One of them intended to represent the same
does it incorrectly. Pick it up. (2018]
velocity distance (A) The average velocity during 0 to3 s |April 4, 2025 (D|
(B) The average velocity during 3 to5 is 10 m/s
(a) + position (C) The instantaneous velocity s is 0 m/s
(b) at ( = 2 s is 5 m/s
(D) The average velocity during 5 to7
s and instantaneous
position
velocity at (=6.5 s are cqual
velocity (E) The average velocity from t
(c) Choose the correct answer from=0 the
to t=9 s is zero
time (d) + time (a) (A), (D), (E)only options given below:
(c) (B), (D), (E) only (b) (B), (C), (D) only
16. A car covers the first
half of the distance between 22. A (d) (B), (C, (E) only
at 40 km/h and other particle moves along the x-axis and
half at 60 km/h. The averagetwo places x varying with timet has its displacement
the car is speed of X= Ct-2) +c(t-2)2 according to the equation
(a) 40n/h (b) 45 kmh |Online May 7, 2012
(c) 48 km/h (d) 60 kmh where cn and c are constants of
Numerice Value Questions Then, which of the following appropriate dimensions.
Aperson (a) the acceleration statements is correct?
(b) the acceleration of the particle is 2c, 1April 3. 2025
17.
travell1ng on a
velocity v, for a distancestraight line
xand with moves with a uniform
a uniform (c) the initial of the particle is 2c
for the next
3 velocity v, (d) the velocity of the particle is 4c
distance. The average acceleration of the particle is 2(c +
50 velocity in this 23. The velocity-time Co)
motion is m/s.If, is 5m/s then line is shown in graph of an object moving along astraight
7 v=
ms. object between t=0figure. What is the distance covered byvthe
18. Ahorse rider covers half the |April 2, 2025 (I)) to t = 4s ? |.Jan. 28. 2025 (1D
distance
The renaining part offthe distance was withS m/s
travelled with speed.
10 m/s for half the time and with speed 15
ms for th speed
of the time. The mean speed of ithe rider
whole time ofmotion is x/7 m/s. The valueaveraged
of xis
over the
|Jan. 30, 2023
()] (a) 30m
b) 10m
{c) !3 rm
 Motion in a Straight Line A19
24. The motion of an aeroplane is represented by velocity-time 33 The nosition of aparticle related to time is given by x=
graph as shown below. The distance covered by
in the first 30.5 second is aeroplane (5t-4t +5) m. The magnitude of velocity of the particle at
km. |Jan. 23, 2025 (I)| |April 15. 2023 (1)
t= 2s will be:
v(m/sec) 4 ms-! (c) 16 ms! (d) 06 ms-!
A (a) 10ms! (b) 14
400 34. The distance travelled by an object in time t is given by s
-(2.5)t². The instantaneous speed of the objectatt =5s will
be: |April 13, 2023 (11D)
200 B (a) 12.5 ms (b) 62.5 ms!
(c) S ms! (d) 25 ms
35. From the v - t graph shown. the ratio of distance to
displacement in 25 s of motion |April 11, 2023 (I))
2 10 20 30 40 t(sec)
3
(a) 9 (b) 6 (c) 3 (a) 20
(d) 12
25. A particle is moving in a straight line. The variation of 10--
position 'x' as a function of time 't' is given as x = (t-6t m/s 20 25
20t + 15) m. The velocity of the body when its acceleration+ (b)
0 10 15
becomes zero is : (Jan. 29, 2024 (I1)) 5 t(s) ’
(a) 6m/s (b) 10m/s (c) 8 m/s (d) 4 m/s (c) 3
26. Two cars are travelling towards each other at speed of 20 m (d) 1 -20
S-each. When the cars are 300 m apart, both the drivers 36. Given below are two statements:
apply brakes and the cars retard at the rate of2 ms2, The Statement I: Area under velocity- time graph gives the
distance between them when they come to rest is : distance travelled by the body in a given time.
|April 9. 2024 (11| Statement II: Area under acceleration- time graph is equal
(a) 200m (b) 50m (c) 100m (d) 25 m to the change in velocity- in the given time.
27. A body travels 102.5 m in nh second and 115.0 m in In the light of givenstatements, choose the correct answer
(n+ 2)th second. The acceleration is : |April 4, 2024 (D1
from the options given below. |April8, 2023 (11))
(a) 9 m/s² (b) 6.25 m/s (c) 12.5 m/s² (d) S m/s (a) Both Statement Iand Statement II are true.
28. The relation between time't' and distance 'x ist= ax'+ Bx, (b) Statement I is correct but Statement II is false.
(c) Statement I is incorrect but Statement II is true.
where a and B are constants. The relation between
(d) Both Statement Iand Statement II are false.
acceleration (a) and velocity (v) is : |Jan. 31, 2024 (1)| 37. Aparticle starts with an initial velocity of 10.Oms along x
(a) a=-2av3 (b) a=-4av direction and accelerates uniformly at the rate of
(c) a=- 3ay2 (d) a=-5av 2.0 ms2, The time taken by the particle to reach the velocity
29. Abody starts moving from rest with constant acceleration of 60.0ms is |April6, 2023 (IL)]
covers displacement S, in first (p - 1) seconds and S, in (a) 6s (b) 3s (c) 30s (d) 25s
first pseconds. The displacement S, +S, will be made in 38. Match Column-I with Column-II :
tme: |Jan. 29, 2024 (1)| Column-I Column-II
(x-t graphs) (v-t graphs)
(a (b) (2p-2p + 1) s A.
(c) (2p +1)s (d) (2p-1)s
30. Abullet is fired intoa fixed target looses one third of its
velocity after travelling 4 cm. It penetrates further
Dx 10m before coming to rest. The value ofD is:
|Jan. 27, 2024 (II)
(a) 2 (b) 5 (c) 3 (d) 4 C
3l. The velocity ofa particle is v =v tgt+f". Ifits position is
X=0 att0, then its displacement after unit time(1=1) is
(20071
(a) Vtg /2 +f (b) Vot2g+3f
(c) 'otg/2+ f/3 D.
(d) Votg+f V
12. Aparticle located atx=0 at time=0, starts moving along
with the positive x-directionwith a velocity'v' that varies as
V= avr.The displacement of the particle varies with time Choose the correct answer from the options given below:
as
[2006| |Jan. 30, 2023 (DI
(a) 2 (b) t (a) A-II, B-IV,C-I, D-I (b) A-I, B-II, C-III, D-[V
(c) 1/2 (d) (c) A-II, B-III, C-IV, D-I (d) A-I, B-III, C-IV,D-II
 A20 Physics
|05 Sep.
39, The distance travelled i by a particle is related to timet as x= 2020 1
41. The velocity of the particle (m's)
att= Ss IS
|Jan. 2S, 2023 (lD) ’(in s)
(a) 40 ms I
(n) 25 ms !
34 6
2
(c) 20 ms ! (d) 8ms !
40. The velocity time graph ofa body moving in a 37
(b) 12m (c) 11 m 49
shown in figure. straight i |Jan. 24. 2023 (4D| (a) (d)
4

48. Abullet of mass 20g has an initial speed of 1 ms

before it starts penetrating a mudi wall of
,20 juE
the wall offers a mean resistance of2.5x10-2 N, the cm.
thickness
the bullet after emerging from the other side of the speed g
(ms

close to: wallik

|10 Apr. 2019
(a) 0.l ms (b) 0.7 ms (c) 0.3 ms (dy)
time (s) 0.4 ms!
Theratio of displacement to distance travelled by the body 49. The position of a particle as a function of time t, 1S
in time 0 to 10sis a)= at + br-c where, a, b and c givenby
are constants.
(a) 1:I (b) 1:4 (c) 1:2 the particle attains zero acceleration, then its When
(d) 1:3
be: velocity wil
41. dx 19 Apr. 2019 11
lf t=v+4, then
is:|July 29, 2022 ())
|dt =4 (a) a (b) a+ (c) a+
(d) a
(a) 4 4c 3c
(b) Zero (c) 8
42. The velocity of the bullet (d) 16 50. Aparticle starts from origin O from rest and moves2 witk.
becomes one third after it
penetrates 4 cm in a wooden block. Assuming that uniform acceleration along the positive X-axis.
facing a constant resistance during its bullet is figures that correctly represents the motion Identify al
The bullet stops completely motion in the block. qualitativelv
=acceleration, v = velocity, x= displacement, a
(4 + x) cm inside the block. after travelling at t= time)
The value of x is: |8 Apr. 2019 1
(a) 2.0 |July 27, 2022 (1D)
(b) 1.0 (c) 0.5 (d) 1.5
43. A small toy starts moving from the
constant acceleration. If it travels a position of rest under a
the distance travelled by the toy in distance of 10 m in t s. (A) a (B) v
the next t s will be:
(a) 10m |June 29, 2022 (D]
(b) 20m
(c) 30m (d) 40m
t
44. Two buses P and Q start
from a point at the same time and
move in a straight line and their
positions are represented
by X)=au t Br and X)=ft-P.
buses have same velocity ? At what time, both the
|June 25, 2022 (11)) (C)
(D)
(a) (b) atf a+f
2(B- 1) (c) 2(1 + P) (d)
t
1+B (a) (B). (C)
45. The instantancous velocity ofa 2(1 +B) (b) (A)
(c) (A),
line is given as v= at t particle moving in a straight 51. Aparticle(B), (C) (d) (A), (B), (D)
The distance travelled by Bt,the where a and B
are constants starts from the origin
particle between Is and 2s is: the positive at time t =) and moves: along
is shown in r-axis. The graph of velocity with hme
|July 25, 2021 (1)}
t Ss? figure. What is the respect totime
(a) 3a+ 7B (b) position of the particle at
1|
|10 Ja. 2019
B 3 7
(m/s)
(c) (d)
2 3 3
46. Ascooter accelerates fromn rest for
time (, at 2
a, and then retards at constant rate a,
for time constant rate
1, and comes
to rest. The corrcct value of will be |Feb. 26, 2021 (1)
(a) 10m 2 3
6
52. In a
car race on(b) 6m
(a) (b) (c) 9 10 t (s)
(d) (c) 3 m (d) 9 m
47. The velocity (v) and time () graph of a
car Bat
the straight
finish road, Car Atakes a time t less
than

body astraight
motion is shown in the figure. The point Sis ain 4.333
'V more
than of car and spe
sat line
seconds. with
passes finishing point withandatrave
B. Both
The total distance covered by the b0dy in constant
6s S: cqual to: acceleration the cars Sstart from rest Then
a, and rroctily
 Motion in aStraight Line A21

(a) 2ad2 (b) 2a, a, t

x(m)
a+ a
(c) Va, a, t (d) a + a,
2
53. An automobile, travelling at 40 kmh, can be stopped at a 10

distance of 40mby applying brakes. If the same automobile

is travelling at 80 km/h, the minimum stopping distance, in
metres, is (assume no skidding) |Online April 15, 201I81 A
(a) 7Sm (b) 160m (c) 100m (d) 150m
Ar 4m
54. The velocity-time graphs of acar
and a scooter are showm in the A Car B 0 10 20 (s)
45 +
figure. (i) the difference between ’(ms)
Velocity

(a) 4=V-0.5 m/s (b) V,= 0.5 m/s< VR

the distance travelled by the car E Scooter,G
i 30 (c) v,=0.5 m/s> vR (d) v,= V=2 m/s
and the scooter in l15 s and (i) the
58. An object, moving with a speed of6.25 m/s, is decelerated at
time at which the car will catchup 15 + d
with the scooter are, respectively arate given by dt
=-2.5Vv where v is the instantancous
|Online April 15, 2018]
5 10 15 20 25
D
speed. The time taken by the object, to come to rest, would
(a) 337.Sm and 25s Time in (s) be: |2011|
(b) 225.5m and 10s (a) 2 s (b) 4s (c) 8s (d) 1s
(c) 112.Sm and 22.5s 59. A body is at rest at x = 0. At t = 0, it starts moving in the
(d) 11.2.5m and 15s positive x-direction with a constant acceleration. At the same
55. Aman in a car at location Q
on a instant another body passes through x = 0 moving in the
positive x-direction with a constant speed. The position of
straight highway is moving with the first body is given by x, () after time ''; and that of the
speed v. He decides to reach apoint d second body by x,(t) after the same time interval. Which of
P in a field at a distance d from
the following graphs correctly describes (x-x,) as afunction
highway(point M) as shown in the Q R
of time r? |20081
figure. Speed of the car in (x-x) (x-x)
the field is half to that on the highway. What should be the
distance RM, so that the time taken to reach P is minimum?
|0nline April 15, 2018| (a) (b
d d
(a) (b) 2 (-x) (K-)
d
(c) 2 (d) d
S6. The distance travelled by a body moving along aline in time (c) (d)
tis proportional to .
The acceleration-time (a, ) graph for the motion of the body 60. A
car, starting from rest, accelerates at the rate fthrough a
will be |Oniine May 12. 2012| distance S, then continues at constant speed for time t and
then decelerates at the rate to come to rest. If the total
2
a distance traversed is I5 S, then (2005)

(a) (b (a) (b) S=ft

61. Aparticle is movingeastwards with a velocity of 5ms. In

10 seconds the velocity changes to 5ms-'northwards. The
(c) (d) average acceleration in this time is (20051
1 -2
(a) towards north
57. The graph of an object's motion (along the x-axis) is shown
in the figure. The instantaneous velocity of the object at (b) -ms towards north - east
points Aand Bare v, and v, respectively. Then
{Online May 7. 2012]
 A22
Physics
such
that its
straight line
73. increasing
Aparticle
in a
moving
isat 5 ms' per velocity
meter. The. acceleration of the particleis
(c) ms
towards north - west where its velocity is 20 ms
ms at a point 1July 25. 2022 (il
(d) Zero
tacceleration 'a'
62. The relation between time t and distance x is
where a and b are constants. The acceleration 1s
t= a 74. graph
is
shows
Aparticle
moving with
constant
v² versus x(displacement) plot. The Following
(2005) m/s².
(a) 2by3 is
(b) -2aby2 (c) 2av (d) -2a3 acceleration of theparticle
63.
11 a body looses halfof its velocity on penetrating 3 cm in a |NA. Aug. 31. 2021 (iL.
wOoden block, then how much will it penetrate more
Coming to rest? bee
|2002|
(a) Icm (b) 2m
Numeric Value Questions
(c) 3m (d) 4cm.
v(m/s 80 -

60
64. Twocars P and O are moving ona road in the same direction. 40+
Acceleration of car P increases linearly with time whereas 20
car moves with a constant acceleration. Both cars cross
cach other at time t= 0. for the frst time. The maximum 10 20 30

possible number of crossing(s) (including the crossing at x (m)

t= 0) is displacement x is piven
65.
(Jan. 29. 2025 (II)|l
75. Ifthevelocity of abody related to
A particle moves in a straight line so that its displacement x by y= y5000+ 24x m/s, then the acceleration of the body
at any time t is given by x² = | +t2. Is acceleration at any |NA, Aug. 27, 2021 (l
time t is xn where n = |Aprii6, 2024 (II)] m/s'.
shown in the
The speed verses time graph tor a particle is
66.
A body moves on a frictionless plane starting from rest. If 76.
S, is distance moved between t=n - l and t=n and figure. The distance travelled (in m) by the particle during
be
S,. distance moved between t=n-2 and t =n-1,then
S,is the timeinterval t= 0 to t=5swill
|NA 4 Sep. 2020(b
the ratio Sl
Sn
is(1-2) for n = 10. The value ofx is. 10T
[April 5, 2024 ())
8t
6
67. A bus moving along a straight highway with speed of 72 (ms ) 4
km/h is brought tohalt within 4s after applying the brakes.
The distance travelled by the bus during this time (Assume 2
the retardation is uniform) is m. [April4. 2024 (I1)I
1 2 3 4 5
68. A particle is moving in one dimension (along x axis) under time
the action of avariable force. It's initial position was 16 m
right of origin. The variation of its position (x) with time (t) (s)
77. The distance x covered bya particle in one dimensional
isgiven as x =-3t + 1812+ 16t, where xis in m andtis in s.
The velocity of the particle when its acceleratiíon becomes motion varies with time t as x' = at' + 2bt + c. If the
zero is m/s. |Feb. 1, 2024 (1)| acceleration of the particle depends on x as x", where nis
69. A particle initially at rest starts moving from reference point an integer, the value ofn is |NA9Jan 2020 1|
x = 0 along x- axis, with velocity v that varies as
V= 4x m/s. The acceleration of the particle is ms2,
(Feb. 1, 2024 (II)] Topic 3-Relative Velocity
70. The displacement and the increase in the velocity of a
moving particle in the time interval oftto (t + 1) sare 125 m
and 50 m/s, respectively. The distance travelled by the
particle in (t + 2)h s is m. (Jan. 30, 2024 () I Multiple Choice Questions
For a train engin moving with speed of 20 ms, the driver 78.
71. 500 m before the station Abody projected vertically upwards with a certain speeu
must apply brakes at a distance of from the top of a tower reaches the ground in t If it is
If the brakes were
for the train to come to rest at the station. projected vertically downwards from
the train
the same speed, it reaches the ground the same point *
applied at. halfof this
distance, engine would cross
with speed Vxms The value of x is to reach the in t,. Time requ
the station [Feb. 1,2023 (1I)| ground, if it is dropped from the top of the
tower, is :
is produced by brakes) |April 6. 2024 (I1!
(ASsuming same retardation
km/h and after appline (a) tit2 (b)
72. A car is moving with spced of 150it stops. Ifthe same car ie ty -t2 (C)
thebreak it will move 27 m before
reported Speed then 79. Train Ais
movingwith a speed of one thirdthe m distance. moving72along towards
travelling north with speed km/htwo train Brail
and parallel tracks towards
is moving
it will stop after |July 25, 2022 (D| south with speed 108 km/h.
Velocity train B with respect
 Motion ina Straight Line A23
to A and velocity of ground with 87. A car is standing 200m behinda bus, which is als0 at rest.
(in ms-):
respcct to B are
|Fcb. 1, 2024 (1D) The two start moving at thc same instant but with different
(a) 50 and-30 (b) -50and 30 forward accelerations. The bus has acccleration 2 m/s² and
(c) -30 and 50 (d) 50 and - 30 the car has acceleration 4 m/s². The car will catch up with the
80. Two trains 'A' and B' of length " and '4/ are bus aftera time of: |Online April 9, 2017)
travelling into
a tunnel of length 'L in parallel tracks
from opposite (a) 110s (b) /120s (c) 10/2s (d) I5s
directions with velocitics 108 kmh and 72 km/h, respectively. 88. A person climbs up a stalled escalator in 60s. Ifstanding on
If train 'A takes 35 s less time than train thesame but escalator running with constant velocity he
'B' to cross the
tunnel then, length L' of tunnel is: takes 40 s. How much time is taken by the person to walk up
(Given L=607) |April 13, 2023 ()| the moving escalator? |Online April 12, 2014|
(a) 1200m (b) 2700m (c) 1800m (d) 900m (a) 37 s (b) 27 s (c) 24 s (d) 45 s
81. Apassenger sitting in a train Amoving at 90 km/h observes 89. Agoods train accelerating uniformly on a straight railway track,
another train B moving in the opposite direction for 8 s. If approaches an electric pole standing on the side of track. Its
the velocity of the train B is 54 km/h, then length of train B engine passes the pole with velocity u and the guard's room
is passes with velocity v. The middle wagon ofthe train passes the
|April 13, 2023 (II)) pole with a velocity. [Online May 19, 2012]
(a) 80m (b) 200m (c) 120m (d) 320m
82. The position-time graphs for two students Aand
from the school to their homes are shown in figure:
Breturning (a) (b) 2
2
|April 10, 2023 (1))
(A) A lives closer to the school
(B) B lives closer to the school
(C) A takes lesser time to reach homne Topic 4-Motion Under Gravity
(D) Atravels faster than B
(E) B travels faster than A
Choose the correct answer from the options given below:
(a) (A) and (E) only (b) (B) and (E) only
I Multiple Choice Questions
(c) (A), (C) and (E) only (d) (A), (C) and (D) only 90. A
ball is thrown vertically upward with an initial velocity of
83. A boy reaches the airport and finds that the escalator is not X +1
150 m/s. The ratio ofvelocity after 3 s and 5 s is The
working. He walks up the stationary escalator in time t,. Ifhe
remains stationary on a moving escalator then the escalator value of x is
takes him up in time t,. The time taken by him to walk up on Take (g = 10m/s). [April 12. 2023 (1)|
the moving escalator will be : |July 20, 2021 (I1)] (a) 6 (b) 5 (C (d) 10
91. A ball is thrown up vertically with a certain velocity so that,
tt2 tËtg
it reaches a maximum height h. Find the ratio of the times in
(a)
tz + t; () t,-t;
(b) 2
(c)
84. Train A and train B are running on parallel tracks in the which it is at height while going up and coming down
3
opposite directions with speeds of 36 km/hour and 72 km/
respectively. July 29, 2022 (1)
hour, respectively. A person is walking in train A in the
direction opposite to its motion with a speed of 1.8 km/hour W2-1 V3+V2
(a) (b)
w.r.t train A. Speed (in ms) ofthis person as observed from V2 +1 V3-V2
train Bwill be close to : (take the distance between the tracks V3-1
as negligible) [Sep.2, 2020 ()] (c)
(a) 29.5 ms-! (b) 28.5 ms-! V3+1
(c) 31.5 ms-! (d) 30.5 ms-!
92. Aball is released from a height h. If t, and t, be the time
85. Apassenger train oflength 60 mtravels at aspeed of 80 km/hr. required to complete first halfand second halfof the distance
Another freight train of length 120mtravels at a speed of 30 respectively. Then, choose the correct relation between t,
and t,. July 29, 2022 (1)
km/h. The ratio of times taken by the passenger train to
completely cross the freight train when: (i)they are moving
insame direction, and (ii) in the opposite directions is:
(9) 4-V2) (b) 4=(V2-I);
|Jan. 12, 2019 II)
11 3 25 93. Ajuggler throws balls vertically upwards with same initial
(a) (b) (c) (d)
5 2 2 11 velocity in air. When the first ball reaches its highest
86. Aperson standing on an open ground hears the sound of a position, he throws the next ball. Assuming the juggler
jet aeroplane, coming from north at an angle 60 with ground throws n balls per second, the maximum height the balls
level. But he finds the aeroplane right vertically above his can reach is
position. Ifv is the speed of sound, speed of the plane is: |Jul 29, 2022 (D)
|Jan. 12, 2019 TI| (a) g/2n (b) g'n (c) 2gn (d) g/2n?
21 94. ANCC parade is going at auniform speed of 9kmh under
(a) (b) (c) v (d) a mango tree on which a monkey is sitting at a height of
2
19.6m. At any particular instant, the monkey dropsa mango.
 A24
101. Theposition, velocity and acoceleration of a particle moving
Physlcs
A Cadet will receive the mango
whose distance from me acceleration can be represented by:
tree at time of drop
(Giveng9,8 m/s)
(a) Sm
is :

(b) 10nm (c) 19.8m

A bullet is shot vertically downwards with an
|July 28. 2022 (1))
(d) 24.5 m
with aconstant

posito velocity acelrtion

|March 18, 20021

a(t)
(

Ot T00 m/s from acertain heicht. Within initial velocIty

10s,
reaches the ground and instantaneously comes totherestbullet
x()
(a)
aue
to the perfectly inelastic coll ision.
for total time (= 20s will be:

(a)
The velocity-time
(Takeg= 10m/s)
curve
1.July 27. 2022 (D|
positon x(t)
velocity v í t )
acelrtion a(t)
(b)
+ 100 m/s
(b)

acelrtion
10 20 s
--100 m/s - 100 m/s
1Os 20s

-200 m/s 200 m/s

positon x(t)
velocity v(t)
a(t)
(c) (c)
(d)
+ 100 m/s +

dcelraton
106
20s 10 s 20s

96,
I00 mis

Two balls A and B are placed at the top

100 m/s

postion x(t)
Velocity v(t) a(t)

(d)
Ball A is released from the top at t = 0 of 180 m tall tower.
s. Ball B is thrown
vertically down with an initial
certain time, both balls meet 100velocity'u' at t= 2 s. After a
m above the ground. Find
the value of'u' in ms.[use g= 10 102. A stone is dropped from the top of a building.
ms]: When it
|June 29, 2022 (DI crosses a point 5 m below the top, another stone starts to
(a) 10 (b) 15 (c) 20 fall from a point 25 m below the top. Both
97. (d) 30 stones reach the
Water droplets are coming from an open tap at a bottom of building simultaneously. The
rate. The spacing between a droplet observed particular height of the
second after its fall to the next droplet is 34.3 m. At at 4th
building is |Feb. 25. 2021 (|1)l
the droplets are coming from the tap? what rate (a) 35m (b) 45m (c) 25 m
103. A helicopter rises from rest (d) 50m
(Take g =9.8 m/s) |June 25, 2022 (1 )) on the ground vertically
(a) 3 drops/2 seconds (b) 2 drops / second upwards with a constant acceleration g. A
dropped food packet is
(c) 1drop /second (d) 1drop /7 seconds from the helicopter when it is at a
98. Water drops are falling from a nozzle of a time taken by the packet height h. The
floor. from a height of9.8 m. The drops fall at ashower onto the the accelertion due to to reach the ground is close to \gs
oftime. When the first drop strikes the floor, at regular interval gravity]: |Sep. 5, 2020 (1))
the third drop begins to fall. Locate the positionthat instant, (a)
drop from the floor when the first drop strikes the offloor.
second (b) t=1.8
|Aug. 27. 2021 (II)) Vg
(a) 4.18m (b) 2.94m (c) 2.45m (d)
99. Water droplets are coming from an open tap at a7.35m (c) t= 2h
rate. The spacing between a droplet observed at 4h particular V3g (d) t=3.4,
second 104. ATennis ball
affer its fall to the next droplet is 34.3 m. At what
rate the is released
droplets are coming from the tap ? (Take g = 9.8 m/s) falling on a wooden from a height h and after freely
|July 25, floor it rebounds and reaches height
(a) 3 drops / 2 seconds (b) 2 drops / second 2021 (1 .The velocity versus
(c) 1drop / second (d) 1 drop /7 seconds may be represented height of the ball during its motion
100. A balloon was moVing upwards with a uniform velocitvof (graph are drawn graphically by:
10m/s. An object of finite mass is dropped from the balonn
when it was at a height of 75m from the ground level
schematically andi on not to scale)
|Sep. 4, 2020
height of the balloon from the ground when object strilo
the ground was around: h2
(takes the value of gas 10 m/s?) (a)
July 25, 2021 (I h2
(a) 300m (b) 200m (c) 125m (d) 250m h h(v) (b) h(v)
 Motion in a Straight Line A25

8
(a) meters from the ground
(c) h(1) (d) 7h
h/2 meters from the ground
(b)

(c) meters from the ground

105, A body is thrown vertically upwards. Which one of the
following graphs correctly represent the velocity vs time? 17h
(d) meters from the ground
|2017| 18
110. Froma building two balls 4and Bare thrown such that Ais
thrown upwards and B downwards (both vertically) with
(a) (b) same speed. If v, and vp are their respective velocities on
reaching the ground, then |20021
(a) Va VA
(b) V= VB
(c) V VB
(d) their velocities depend on their masses.
(c) (d) Nunerie Value Questions
111. A body falling under gravity covers two points A and B
106. Two stones are thrown up simultaneously from the edge of separated by 80min 2s. The distance of upper point Afrom
a cliff 240 m high with initial speed of 10 m/s and 40 m/s the starting point is m (use g=10 ms2)
respectively. Which of the following graph best represents Jan. 27, 2024 (ID)
the time variation of relative position of the second stone 12. Atennis ball is dropped on to the floor from aheight of9.8
with respect to the first ? m. It rebounds to a height 5.0 m. Ball comes in contact with
(Assume stones do not rebound after hitting the ground the floor for 0.2s. The average acceleration during contact
and neglect air resistance,takeg- 10m/ s) is ms². [Given g= 10 ms] |Jan. 29, 2023 (I)|
(The figures are schematic and not drawn to scale) |2015) 113. Aball is thrown vertically upwards with a velocity of 19.6
ms- from the top of a tower. The ball strikes the ground
(a)
240Ty:-y)m
(b)-y)m
240 after 6s. The height from the gound up to which the ball can
rise will be m. The value ofk is (use
8 12>(s) 1>s) g=9.8 m/s) [July 28, 2022 (I)
114. Aball of mass 0.5 kg is dropped from the height of 1Om. The
0,-y)m
240 an 2-y) m height, at which the magnitude of velocity becomes equal to
(c) the magnitude of acceleration due to gravity, is m.
(d)
(Use g = 10 m/s²). |June 26. 2022 (1)|
[’ 8 12 >1(s) 2 ’ (s) 115. Aball is projected vertically upward with an initial velocity
of 50 ms at = Os. At t = 2s. Another ball is projected
107. From a tower of height H, a particle is thrown vertically vertically upward with same velocity. At t= S.
upwards with a speed u. The time taken by the particle, to second ballwill meet the first ball(g = 10 ms).
hit the ground, is n times that taken by it to reach the highest
|June 26, 2022 (D)
point of itspath. The relation between H, u and n is: (2014| 116. From the top of atower, a ball is thrown vertically upward
(a) 2gH=nu? (b) gH= (n-2) u²d
(c) 2gH= nu² (n-2) (d) gH= (n -2)u2 which reaches the ground in 6s. A second ball thrown
108. Aparachutist after bailing out falls 50 mwithout friction. vertically downward from the same position with the same
When parachute opens, it decelerates at 2 m/s . He reaches speed reaches the ground in 1.5 s. A
third ball released, from
the ground with aspeed of3 m/s. At what height, did he bail the rest from the same location, will reach the ground in
out ? |20051 S
|June 24. 2022 D
(a) 182m (b) 9lm i17. Aball is dropped from the top ofa 100 m high tower on a
(c) 11lm (d) 243m
1
109, Aball is released from the top of atower of height h meters. planet. In the last ,sbefore hitting the ground, it covers a
It takes Tseconds to reach the ground. What is the position distance of 19 m. Acceleration due to gravity (in ms ) near
T
of the ball at second |2004| the surface on that planet is (N4 Jan. 8. 2020 1
3
 Physics
A26
P a r t - B ( J E E A d v a n c e d )

acceleration-displacement

most
suitable
graph wil
The a
bo

Topic 1-Distance, Displacement & (b)

Uniformn Motion
(a)

1 MCQs with One Correct Answer

movingin a
point B,
1.
In 1.0 s, a particle goes from point AtoThe magnitude ofthe
semicircle of radius 1.0 m((see Figure). -2 |19995
Marks]
(d)
average velocity
(c)
(a) 3.14 m/s
(b) 2.0 m/s timet= 0, the acceleration
rest at time
1 Om
starts from
(c) 1.0m/s 6. Abody
shown in the figure.
The maximum velocity atained
graphis 1200481
will be
(d) Zero by the body Acceleration
(a) 110m/s (m's') 10
Fll inthe Blanks
radius R. In half the period oI
2. A particle moves in a circle of and distance (b) 55 m/s
revolution its displacement is |1983 -2 Marks)
650m/s
covered is (c) Time
(sec.)
More than One Correct Answer (d) 550 m/s
6 MCQS with One or
with a velocity of 5 m/s. In Integer Value Answer
3 A particle is moving eastwardsm/s northwards. The average constant
10s the velocity changes to 5 gravity free space with a
Arocket is moving in a
Marks [1982 - 3
acceleration in this time is/are 7. direction (see figure). The
acceleration of 2 m/s along +x ball is throm
(a) zero rocket is 4 m. A
1/V2 m/s² towards north-west length of achamber inside the
(b)
from the left end of the chamber in +x direction with aspeed
1/J2 m/s² towards north-east the same time, another hal
(c) of0.3 m/s relative to the rocket. At
of0.2 m/s from its right
(d)
1 mls towards north-west
is thrown in -x direction with a speed when the two
time in seconds
2 end relative to the rocket. The |Ady. 201
(e) 1
m/s2,towards north balls hit each other is
2 m/s
0.3 m/s 0.2 m/s a=2

4.
10 Subjective Problems

Answer the following giving

Time> 4m
X

reasons in brief:
Correet Answer
Is the time
variation of 6 MCQs with One or More than One
follows : it starts
position, shown in the figure 8 Aparticle of mass m moves on the x-axis as
observed in nature? |1979] Position (x)’ from rest at t= 0 from the point x=0., and comnes to rest at 1=1
at the point x= 1. NO other information is available about is
motion at intermediate times (0 < t ). If a denotes e
Marks
instantaneous acceleration of the particle, then: |1993-2
Topic 2-Non-uniform Motion 0<IS.
(a) acannot remain positive for all tin the interval
(b) a cannot exceed 2 at any point in its path.
(c) a must be 4 at some point or points in its path.
Correct Answer no other
1 MCQ: With One (d) a must change sign during the motion, but
velocity-displacement graph of a particle moving assertion can be made with the information giVel.
The
5. straight line is shown |2005S1
along a
10 Subjective Poblens
9 sometime
Acar
accelerates from rest at a constant rate oafor to rest.

after which it decelerates at a comet

constant rate Bto
If the total time lapse is t seconds. evaluate.
(1978)

maximum velocity
(ii) the total distance reached, and
travelled.
 Motion In a Straight Line A27

move in the oppositedirection to thetrain, while the distant

objects appcar to be stationary
Topic 3-Relative Velocity in STATEMENTT2:Ifthe observer and the object are moving
One Dimension at velocities v, and V, respectively with reference to a
laboratory frame, the velocity of the object with respect to
the observer is V, - V: |200%;
1 MCQS with One Correct Answer (a) Statement-1 is True, Statement-2 is True, Staternent-2
10. Airplanes Aand Bare flying with constant velocity in the is acorrect explanation for Statement
same vertical plane at angles 30° and 60 with respect to the (b) Statementl is True,Statement-2 is True: Statement-2 is
NOT a correct explanation for Statement- 1
horizontal respectively as shown in figure. The speed of 4
is 100/3 m/s. At time t =0 s, an observer in A finds Bat a (c) Statement -1 is True, Statement-2 is False
distance of 500 m. The observer sees B moving with a (d) Statement -1 is False, Statement-2 is True
constant velocity perpendicular to the line of motion of4. If 10 Subjective Problens
at t= Ajust escapes being hit by B, t, in seconds is 14. On a frictionless horizontal surface, assumed to be the x-y
| Adv. 2014] plane, a smalltrolley Ais moving along astraight line parallel
4 to the y-axis (see figure) with a constant velocity of
(V3-1) m/s. At aparticular in stant, when the line 04 makes
an angle of 45° with the x-axis, a ball is thrown along the
B surface from the origin 0. Its velocity makes an angle o with
the x-axis and it hits the trolley.
.o30
560° (a) The motion of the ball is
observed from the frame of
Fillin the Blanks the trolley. Calculate the
angle made bythe velocity
11. Four persons K, L, M, Nare initially at the four corners of a vector of the ball with the x
square of side d. Each person now moves with a uniform axis in this frame.
speed vin such a way that Kalways moves directly towards (b) Find the speed of the ball 45°

L, Ldirectly towards M, Mdirectly towards N, and Ndirectly with respect to the surface,
towardsK. The four persons will meet at a time ifo=40/3. 12002 - 5 Marksl

|1984- 2 Marks)

True/False Topic 4- Motion Under Gravity

Two identical trains are moing on rails along the equator
on the earth in opposite directions with the same speed.
They willexert the same pressure on the rails. True /False
|1985 - 3 Marks| 15. Two balls ofdifferent masses are thrown vertically upwards
with the same speed. They pass through the point of
projection in their downward motion with the same speed
9 Assertion and Reason/ Statenent Type (Questions (Neglect air resistance). |1983 -2 Marks
13. STATEMENT-I: For an observer looking out through the
window ofa fast moving train, the nearby objects appear to
 Chapter

Motion in a Straight
Line
1. (a) For option
(A) Part-A (JEE Main)
=kt + c
6.
For SHM (ID) (c) Average velocity.
x=A sin#=A sin (kt+ c) Total Displacement X+X 2x + CD=
Average speed of the 3x’CD= 3r-2r=r
V

For option (B) Total time taken Total distance

object <v>
constant 2x Total time
X
dy X X t= and t, = <V>=
3x
2y +2x=0 V;
dx
a=-x
v= 11. ()
So, it is 1D motion
For option (C) 7
()
Time can't be
possible.
negative, so motion is not
A
=0
B
Average velocity V2
For option (D)
Distance, s= f(t), so it can be lD Total displacement 2x
motion. Total time I=3sec
t, +t
2. (d) Total round =1+ 2x I`=Nr'+r =v2r=2x10 =10V2
Displacement =2r 2 X X
2v2 12. (d) Given, First distance m
S, =4 km travelled,
Distance - 2r + 2r =3rr Second distance travelled, S, =4
2 km
R
(a) PQ= RN2 Distance S,
3. X V=
(d) time
P
QS=x=9t+ 15t =24t Here, V, and V, are the velocities during
PQ=X=6t, = 24t ’t,=4t R P
and second distance travelled first
<speed> =
Total distance 48t ’Vay =
4+4 15
Total time 4 4’Vay ==3.75 kmh
21 +t,
48t 48t Let v be instantaneous velocity 3 5
=8m/s 13. (b) From AAOB
2t +4t 6t R AB
2 OB
4. (b) Here, t=
Arc length (PQ) 4 TR
sin 135 sin 22.5
21
Average velocity. AB
sin 135
sin 22.5-OB
<V»isplacement
S,=vx 3t Time
sin (135) arc (AB) sin (45) 60 x 4
80 x t sin (22.5) 37
+80 ×3t sin (22.5)
Yaxg
S, + S 2 PQ_ RN2x2v 2V2v 4
T 4t
=47 m
t TR
=70 km/hr 14. (b) Given, v= bx Or
5. (b d
xV2 2/5X=2
d
P
Displacement 2R cos -0
or
2 = br
9. 2zR
0 W2
R=2 km At
3y
4
180-120 Differentiating w. r. t. time, we get
3x 2R cos
2
R -3 cos 30 dx b x2
2rR d 4
Or y= )(*)

=1.5V3m/s 15. (b) Graphs in option (c) position-time

and option (a) velocity-position are
10. (b) Consider, AB=x corresponding to velocity- time graph option (d)
Displacement of cyclist =PS and its distance-time graph is as given below.
.: Displacement D
Hence distance-time graph option (b) is
A B C incorect.
BC=x
A280

tistave
A (2ux +p)
2a

2ay
Physies
d 29. (a) Using s
slope ut

time verifyig cach nd every option

50
16. (A) 0to 3 scc; V n/s
(c) Average speed
Total distance travelled 55 ap
Total time taken (B) 0to S scc: <V
Displacement
48 kmh (C) 2; slope S m/s
,allp
)-5
2x 40 2x60 2.5 m/s
(D) | 5to 7 sec: <
17.
avg
(10) Average velocity,
tx, AtI 6.5 sec; ý= 10 30. (None) Using v² u?
J2p'-2p+)
(E) I 0to 9: V>=0) After travelling 4cm bullet 2aS
its velocity losses one third
22. (d) For particle moving along x-axis,
2
3x xCo(-2) +c(t 2) 2u
2 50 5/2
Velocity, v
dx
2tc, 2e(t-2) 3 u'2( a)(4× 10)
3x 3 dt
Acceleration, 4u2

7
S 2v,
dt 2c, t 2e 2(c, tc) 9
u
2a(4.10 )
3 7 1 7-4 Su
23. (a) The area under the velocity-time
2v2 20 2v, 20 20 graph gives distance. 2a(4z10 ')
3 When bullet comes to rest,
Or, 2v, 20 V,= 10m/s Distance= -x2 x10+2 x 10 30m

X meters X meters 24. (d) The area under velocity-time graph +2(-a)(x)
18. (50) gives displacement
A B 4u
-2ax
In motion AB, tAR S
2
x(200+ 400) ×2 +28,5 x400 9
Dividing eq. (i) by (ii)
In motion BC, x =d, td, -12000m =12km 4x10 2 16
where d, &d, we the distance travelled with 10 s 25. (c) Displacement, x =t'-6t' +20t+ 15 x=x10
4 X
and 15n's respectively in cqual time intervals .:. Velocity, v
dx
3t² 12t +20 .. x=3.2 x 10 m-32 x10 m-D< om
cach dt
dv
(given)
I0t 15t .. Acceleration, a= 6t- 12 .. D=32
,d, = dt
d, When a =0 31. (c) We know that, y =
dx dr=vd
2x dt
d, td, - x(10 + 1S) = 2
6t-12 =0 >t=2s
2x 2x 25 50
m/s
Att=2s, v= 3(2)- 12(2) +20 - 8 m/s
26. (c) Given, speed of car, u=20 ms Integrating. fd-fvd 0
2x 5+2 For retardation of car, a=-2 ms 2
25 Using v² =u²+ 2aS 2 3
19. (3) >(0= (20) + 2-2)(S) 0

200+ 200 +
2nx 200 S= 100 m
Syotal. = 4 Distance travelled by twocar -2 x 100 -200 m 2 3

Tjotdl 238 Remaining distance =300 -200 =100 m Att= I, x= v) t 2 3

400 + 100x 714 27. (b) Since S, = u+
238 238
3m/s (2n-I)
a 32. (a) v=ur,
Given, 102.5 =ut(2n-) 2 dx
=adt
20. (18) Vavg and 115 =ut(2n+ 3)
Iiotal
3vy Integrating both sides.
1 ’102.5 =u+ an-- & 115= u+an +3a
= alth
+
.. 12.5 2a ’a6.25 m/s?
33 28. (a) Given
198 198 t r+ Br - a t

33 .18 m/s.
dt 4
6+3+2 dx
= 2ax + ß ’ v 33. (c) X=5 41 +5
dx dt 2ax +ß V T 0 t - 4 : Att=2s
y= 16s
13
displacement-time s di, s)r
21. (d) The slope of 20
(x - ) graph gives velocity of object. dx 2ux + B2ux +B 34. (d) Specd (v) (::s-25it'g
Motion in a StraightLine A281

v=St 43. (c) In sec. S= 10 m 49. (b) x at + br- c

At=5.V= 5t =5x 5 25 m/s dx
So, 10 .(i) Velocity, v= Clat +br'-t')
35. (c) Distance area under y dt d
with positive value
-tgraph In next '' sec, suppose body travel S distance -a+ 2bt 3ct
dv d
Then, 10 +s' = ..() Acceleration,
-(a + 2b1 - 3ct')
Distance x10 ×5+ 5 x10 +5 x 10 d dt
Dividing (i) by (i), we get
10 +S =4 ’ 10+S'= 40 ’ S'= 30m or 0=2h-3c x 2I =
+-x5x10+, x5x20+ x 5x 20= 250 m
10
Displacement =arca of graph with sign 44. (d) For bus P
=(200--50)m =150 m and v= a +26| 3c
3c
*,()= at +Br? there is
Distance 250 50. (d) For constant acceleration,
150 straight
Displacement V,(t) = a+2ß1 | : d
6 (a) Area under velocity time graph gives line parallelto t-axis on a-l graph.
For bus 0
distance of body in given time.
x()=ft-t Inclined straight line on v - graph, and pa
V dxvdt
dt
=Area rabola onx-graph.
Area under acceleration time graph gives dt 51. (d) Position of theparticle,
change in velocity in the given time. As, V,() =V() ’a+ 2Br =f-21 S= area under graph (time t=0 to 5s)
dv ’a-f=-2ßr-2t >f-a=2ß1 + 21
’v=adt =Area -x2x2+2x2+3xl=9m
37. (d) Initial velocity, u = 10 ms-! f-a
52. (c) Let time taken by car A to reach
Finalvelocity, v =60 ms1 (2ß + 2)
45. (b) We have given, finishing point is to
Acceleration, a = 2 ms2 V= at+ Bt .:. Time taken by car B to reach finishing point
From the equation of motion, S
ds
y=utat’ 60 = 10 +2t >t= 25 sec.
38. (a) dt
S; 1
dx
(A) dt
= slope >0 always increasing u=0 VA =a,t
dx
(B) dx <0;and at t -’ o ’ 0 Vg a,( +t)
dt As particle is moving in a straight line, VA -YBv
dx dx .:. Distance= Displacement
>0 for first half< 0 for second half. ’V=a, o- a, (% +t) =(a, - a,a,t ...)

(D)
d
dy
dt
= Constant
dt
Distance [al4-],p8-J
2 2 3
1

46. (c) For time interval t,

39. (a) Given, x = 41 »Vat,=ya. (t, +t)
dx For time interval t,
V= -8t
V, = u, t (-a,l,) or, 0 =v -asl,
The velocity of the particle at t = 5s is ya,t
y=8x5=40 m/s.
40. (d) Displacement 4;l a,l, or, 1
= )area =16-8+ 16-8= 16m Putting this value of t in equation ()
v=(a1-a,) vajya,t
Distance = Slarea = 48 m 47. (a) 4 B
-azt
displacement 16m (ms) 2 S -ya
So, D (in s)
Distance 48m O|1 2 3
41. (b) Given, (= x+4 =
Jaja,t ta,t-azt or. v=yaja,t
=(1-4=-8/+ 16 5
2s=4+-Ps:
OS
3
SD 2-
= 2i-8 = - 2x4-8 =0 53. (b) Here, s =
dt Distance covered by the body = area of v-t 2a
graph
42. (c) Let initial velocity of bullet =u = area (OABS) + area (SCD) 80
2

Final velocity of bullet = 37 40

3 m
Distance travelled by bullet, s =4 cm 33 3 DS, = 160 m
Using, v²- u=2as 48. (b) From the third equation of motion
y-u= 2aS 54. (c) Using equation, a = and
&u2 u
=u-2a(4) ’ a= F
9(8) But, a = S=ut +
Using v²-u =2as again m
0 -2a(4 + x) Distance travelled by car in 15 sec
2.5x10 20
ar=()-0x10 |100 1(45)
2 I5 (15) =
675
m
Distance travelled by scooter in 15 seconds
4.5=4+x x=0.5 -m/s =0.7m/s =30 x 15= 450 m
(: distance = speed xtime)
 d) t
dtS - 2 a r y +hy= 2ar byl
A282
Dlerence
betwen
distance
travellcd
3375- |12.5
by car
m
Vvelrca
ISsc. 450
2ax +h=

and scooer in
scooter in time t:
Letcar catches Again)diferentiating.
we get
675 =>
3375+46675-300
1 dv
+ 45-1S) =30r spced
dr +0= -

22.5scc with
constant
2 dt
>1Êt-3R7,5 at a
of the highway
moving
turn
- , =0 d
car body
Let the =N Forthe
55, (a) point M. So, RM 0, x, -2ay'
distance 't' from thc time taken
ficld is v, then OM- 0n
at (= ’ Acceleration

dt
Andif specd of car in =
= a -
d
distance OR -)
In first case:
cover the . 63. (a)
by the car to ofparabola.

the highway, is
This
equation
= , S = 3cm, a, =9
QM -x
..)
slope is
negative
u, =u;V
- ;the
For /<
21
taken to travel the
distance 'RP' in
the
is zero
Using, v -u = 2a;s;
Time slope
- ;the 2
field For = 2xax3 ’a=
. () positive 8
slope is graph (b).
For (> a ;the represented by In second case:: Assumingthessame retardat
to move the car from Qto P characteristics
are rest and
Total timc elapsed These starts from
A from
4,
u, = u/2;V, =0;s, =?;
car
(d) Let
acceleration

QM-x 60. point B with

1 moves up to
2v
Distance, AB= S=
v-u =2a, xs;
Distance, BC- (fi,)
-u xS) ’S,=Icm
.:. 0 4
-,'=2s
).
Distance, CD= 2(f12) 64. (3) For two cars P and
2a constant
’M
C fl24D a, = kt, k is
d B =a, a is constant
For' to be minimum d =0 211
Case-I : If upand a,opin same directioa
d d
= 0or x=F 15 S
J2-1 AD =AB + BC+
CD=15S
Total distance, P
Distance along a line i.e., ’ S+ftt+ 2S =
15S
56. (b) ........ .(1)
’ fit= 12S
displacement (s) = (:s<rgiven) a-kt
ofdisplacement, we get 1 .(ii)
By double differentiation
acceleration.

ds di³ dv d(3') Dividing (i) by (ii), we get i = 6

t
Total number of croSsing =2
Case-II: Ifuop and a in opposite diraa
= 6t
=31 and a= dt
dt dt dt
a=61 0r a ct
Hence graph (b) is correct. 72 P
Ax V
61. (c)
57. (a) Instantaneous velocity y = A
akt
4m
Av =vt(-y)
=0.5 m/s
From graph, V, = 8s
)09

8m
=0.5 m/s
and v = 16s Total number of crossing 3
1.e.,v, =v-0.5 m/s Maximum possible number
of rN
65. (3) Displacement at time tIS
d dv
58. (a) Given,
d -2.5/vT =-2.5 dt Initialvelocity, v =Si, Final velocity, Vy =5j. x²=| +
Integrating, dv=-2.s[dt
Jo
Change in velocity Av =(v, -v) On di fferentiation, 2x
dx - y =!

dt
= i+vË+2yi, cos 90 =5? +5 +0 dv
--2.5|, -2(6.25)' =-2.5t =S/2ms Again on differentiation, x d dt
6.25

-2x 2.5 -2.51 =>-2s [As| M|=|vy|=5 m/s] X.aty=|

59. (b) For the body starting from rest, Av
distance travelled (r) is given by Avg. acceleration :
10 X
Also tan 0= Acceleration is constant
2 66. (19)
which means is in
(towards north-west)the second quadrant.
 Motion in a Straight Line
Body starts from rest .. u 0 A283
d
73.
-I)-}1
[2n-1) 2n-1 .X 2n-1
(100)
ds
=5 msI Height of body when it is dropped from the
tower,
S, dv
So, a = V =20x 5=100 m/ sec
10 >x= 19 ds H
For n
(40) Given initial velocity 74. (1) As y=u+ 2ax
At point'A' Substitute "u" and "H" in cquation ()
12 kmh= 20 m/s, v=0, t =4s 40 +2a.10
wutat =>0=20+
a x4 At point 'B' ..(1)
8 - 5 m s
60= +2a.20
Lsing, v - u =2as Substracting (ii) from (i), we get ..(1) t=t,,t=/t
a=lm/s?
0-20-2-5).s .. S=40 m 79. (b) V,- 20 m/s and V,- 30 m/s
68. (52) Given, X=-33.+ 18t+ 16t 75. (12) Acceleration equals to d
a =1 Velocity of train Bwrt train A.,
On difterentiation vd d
a=
&v= 5000 +24x VBA-Va-V 30- 20 -50 m/s
d
OX-9t +36t + 16 dy
Velocity of ground wrt train B.
dt 12 VGR=Vc-V,=0-(-30) =30 m/s
x 24 =
d 2/5000 + 24x
Again on differentiation Ss000 +24x 80. (c) Time, t=
distance
dv velocity
dv
=-18t+ 36 dx 600+ 4 612 1050
dt =35 l
12 20 30 35
Ifa =0,t=2s =/5000 +24x x = 12m/s2
Vs000 + 24x Total length of tunnelgiven,
Velocity att= 2s 76. (20) 1050
u=-9(2)+ 36 x 2+ 16=52 m/s. L= 60P x 60 = 1800 m
35
69. (8) Relation between acceleration and 81. (d) Velocityof train Bw.r.t.
selocity is:
dv 1 train A=VB -VA
a=v=4Vxx4x. - 8m/s2
dx =54-(-90) = 144 km/h
"0. (175) Considering acceleration is
Distance travelled =Area of speed-time graph 144 x 5
= 40 m/s
CONstant
=,x5x8= 20 m 18
=utat

Time of crossing = length of train

77. (3) Distance X varies with time tas
4+50=uta>a=50 m/s
x= at+ 2bt +c relative velocity
dx
S,4 =ut1n-1] ’ 2x
dt
- 2at + 2b ’ (8) = 40
Displacement in (t + 1lyh second dx (at +b) .:. l=8x40 =320m
= at +b ar
dt dt 82. (a) As slope of B > Slope of A
125 =u++1) d'x + :. Speed ofB (V,) >Speed ofA (VA)
50 =a
125 = u+ dt Also, xp >x ata given time
83. (c) Let L be the length of escalator.
125 = u + 50t + 25 a
dt ) Speed of man w.r.t escalator = = V,
100= u+ 50t ....) ti
d1 X

S-2 =u+2(1+
2 2) 1] ar2
-(at +b)' ac-b? Speed of escalator ==V,
th

Si+2 =u+25(2 t+ 3) Time taken when escalator is moving and man

’ ao« x3 Hence, n=3 is also walking on it
Using equation (i) 78. (a) Height of body when projected L
=t=
.S,2 =u+ 501+ 75 = 175m vertically upwards,
1. (200) We have y²=u' 2as 1
Ify= 0, then s = 500m -H-ut,-gi t L tË t
So, 0=20+2a x 500 Height of body when projected vertically 84. (a) According toquestion, train Aand
>1000a =-400 ’a=-0.4 m/s downwards, Bare running on parallel tracks in the opposite
Ifs =250 m, then direction.
...(ü) 36 kmh
v'= 20 +2x-0.4x 250 H-ut, + ;gý 1.8 km h
4
and (ii),
v'= 400-200 > y²=200 On comparing equation (i)
V=y200 m/s ,=36 km/h =10 m/s
(3) If speed becomes 'n' times by
keeping a' constant, then stopping distance
becomes n² times 72 km/h
B

Here, V’V/3 u(t,+t,)= ;slí-5)-, tt)(4,-)

So, S’ 27 V=-72 kmh -20 m's:
-m i.e. 3 m
u=,g, -,)
 betwecn v andi So the curve
×5= 100 line with 100 as intercept
A284 v(t 5s)- 150- 10
for 10 to 20 sec, v=( 0, so option (A
V(t =3s) 120
X+|

V-1.8kuh 0.5 ms
96. (d) Let us suppose both
V(t=5s) 100 below ball rme
man, 8 man, 4
*man.4t-0.5+
8
10- (-20) Maximumn height,
Isec at high h
91. (b)
For 'A', h = g - 8 0 - 2 10
0.5 + 30- 29.5 m's. height
(60 +120) 1000
km h= [:: Atmaximum =16 > 4 sec
85. (a) = 80-30
2g For ball B
v=0. so, 0 = u - 2 g h ]
11
km
(60+ 120) 1000 u=/2gh h=ult - 2) ,8t-2)2
5
80+ 30
S= ut +-at
86.
As, h/3 b 80 =-2u -G82
80 =-2u - 20 ’ 2u =60
. u= 30 m/s
97. (c) For Ist droplet,
R(Observer)
2 equation.
Distance, PQ= P( Distance =speed xtime) Let t, and t, be two
roots of above hË gt=9.8«(4) =78 4m
Distance, QR =v.t V2gh +,2gh-4xx For 2nd droplet, h, =(78.4- -343) m-
2 3
PQ
cos 60° =
QR v.t
Then.
V2gh-2gh - 4x&x
2 3 and l2
2h, 2x4.1 =3s
9.8
87. (c) 2m/sec²
>4 m/sec Time intered, At =(4-3)s =]s
Car
Bus
J2gh t 4gh3 Rate of droplets fall,
200 m
4gh 3-V2
ar=4m/s', ap -2 m/s V2gh R= =ldroplet/s
Given, u =ug=0, arp=2m/sec V3 At
hence relative acceleration, motion,
92. (d) From equation of 1
Now, we know,
1
S= ut +
98. () Using H=gr ’ 2Hg=
s=ut +at'200=x
2
2t [: u= 0] 9.8x2 =?t= v2s
with the bus after For first ...) 9.8
Hence, the car will catch up 22
time For total height h, For first drop, t =0; For second drop. I =y
t=10/2 second For thirddrop, t =2At
..(ü) Let h be the distance travelled by second er
88. (c) Person's speed walking only is 60 Divide equation (ii) by (i)we have when third drop begins to fall
(lis length of escalator)
Standing the escalator without walking
the 1 .2At =N2»A =
2 (t;+t,)
speed is 40 1
Walking with the escalator going.
51 1 9.8
The person's speed is 60 40 120 --x9.8x =2.45m
V2-1 2 4
120 Height of second drop from floor,
So,the time to go up the escalator t : =24 93. () Time taken by ball to reach highest H-h=9.8- 2.45 =7.35 m
5
second.
point, t = 99. (c)
89. (d) Let 'S be the distance between two 100. (c) Given,
ends 'a' be the constant acceleration Frequency of throw, n= -=
1
T_v=...() Initial velocity of hotair balloon,
u=10ms
The maximum height the balls can reach. For stone
As we know y² - = 2aS or, aS =
2 Using,
Let v, be velocity at mid point.
Therefore, v -u'=24 Hmax from equation (i) v= |Oms

2
2g 2g 2n
94. (a) From h= ut +
1 2h
2
Time taken by mango, 75m

t =
2h 2x 19.6 =2s
90. (b) Given 9.8
Initialvelocity of ball, u= 150 m/s
g10 m/s?
Distance x= vt =x2=5 m
2 h= ut
Using 95. (a) at t=0, u= 100 m/s downwards
for 0 to 10sec 1
.:. V=u-gt ’75 = -10tt+g r
:. v(t=3s) = 150 - 10 x 3 =120 V=u-gt=-100 10 x 10=-200 m/s 2
As v = -100 - gt is straight line equation ’75=-10t +5t
 Motion in a Straight Line A285
'-2- 15=0’t=5 sec.
Height of balloon 104. (c) For uniformly acceleratcd/ 108. () Initial velocity of parachute
H=t75>H= 10 x5 +75 125 m. deaccelerated motion : after bailing out,
101. (a) As v ut at and a constant =t2gh 50 m
So, v- graph will be st. line. Ascquation is quadratic, so, v-h graph will be a
d-t graph will be parallel to -axis parabola u 2x9.8x50 - 14/5
a= -2 m/s
The velocity at ground.
and as x=,+ ut + v=3m/s
at and a constant
at -0. h -d

So, x- graph will be parabolic. 1’2:V increases downwards

- 3-980 3 m/s
d
102. (b) velocity changes its direction 2x(-2) 4
2-’ 3:V decreases upwards
collision 243 n
Top of Building takes
4=0 place 109. (a) We have s = ut +
Sm Initially velocity is downwards (-ve) and then 1
after collision it reverses its direction with lesser ’ h=0x T+-gh
Stone-1 magnitude, i.e. velocity is upwards (+ve).
u,= 10 m/s Notethat time =0 corresponds to the point on Vertical distance moved in time
3
25 m
20 m (from u-u-2g x 5) the graph where h= d. h
105. (a) For a body thrown vertically upwards
2 9
Stone-2 acceleration remains constant (a =-g) and
velocity at anytime t is given by V=u -gt :. Position of ball from ground =h 8h
During rise velocity decreases linearly and 9 9
during fall velocity increases linearly and 110. (b) Ball Ais thrown upwards with velocity
Bottom of Building direction isopposite to cach other. u from the building. During its downward
Hence graph (a) correctly depicts velocity journey when it comes back to the point of
versus time. throw, its speed is equal to the speed of throw
Height of the building = h + 25 (u). So, for the journey of both the balls from
106. (a) For Ist stone, -240 = 10t - point Ato B.We can apply v-u'= 2gh. As u,
For stone-1, 20+h= 10r+g ...(i) g, hare same for both the balls, hence, v, =Vg
’ 5t- 10t-240 =0’t= 8s
I,2
And for stone-2, h=g 1
..(i) For 2nd stone, -240 = 40t
Putting value ofh from eq. (ii) in eq. () st40t-240 t=12s
.. For ts 85, y, -y, =(40-10)t =30t
20+=10*g ..t=2s (Straight line)
For t>8s, y,-y, =240+40t St² (Parabola)
107. (c) Let t be time to reach maximum BI
Therefore, h=;sx10x2? =20 m height. Then 'nt is total time offlight.
.:. Height of the building t=t 111. (45) Using S = ut
=h+ 25 = 20 +25=45 m.
So, h=ut-g" From A’B
103. (c) For upward motion ofhelicopter, t=0
1 O
Starting Point
v'=u'+2gh ’ y=0+2gh ’v= 2gh and H+h=8m-1 H -80 =-v1t --x10r? u=0

Now, packet will start moving under gravity. 80= -2v| 7x10x 2
Let 'r be the time taken by the food packet to H+ut -
80 m
reach the ground. -80 =-2v -20
12 -60 =-2v,
1 1 B
=t+at
S=1 -h=2ght .. v,=30 m/s
From O to A
1
;s-2ght-h=0 H+u-g=g| v=u+2gS
30² =0+2x (10)(S)’900= 20 S
2gh t2gh +4xxh gu gu :S=45 m
112. (120) Given height from ball is dropped, h
2x
g' 2g? =9.8 m

H=
u Ynitial2gh 2x l0x9.8 = 14 m/s
-n(n-2)
2g+N2) 2g (downward direction)
Height to which ball rebounds, h = Sm
’ 2gH =u-n(n -2) v; 2gh - 2xl0x5
=10m/s (upward direction)
 A286
2
10 (135- 180) 2 2x45 Physics
50:-gr-50(-2) -9’-3s. 10
Av |1Om/s+ 14m/s 117. (08.00) Let the ball takes
= 20= 4/-4
24
At 0.2
0=-100 -
gf-(-2)] the ground
imello T
=120 m/s 24= 41 I=6 sec. we Using, S=ut
0.2 direction as tve, g S=0xl
Taking upward
116. (3) 200 = gt
19.6
113. (392) t, =2s: v=u- gt get (:25
9.8
h=ux 6
;gx6 Case I 200
0= u-gt
,(6-2)s = 2ha =ug
h=-ux
L5-gx1.5’ Case II

16x 9.8 392 k In lasts, body travels a distance

2
m=
m k=392 h=
’Case III of 19 m.
1
114. (5) =r2gh g6'--1 Su I-distance
2 travelled =81
g-+ 2gh' 0 So, bu
(from I and II]
10 m g(33.75)
1
7.5u=-g (6-1.5): u= 2x7.5
Now. -81 g
So, req. height
= 10-h'= 10 -
45
m s.
81x2
1
=5 m
Again, g-6u-;g6 (from I andIII]
(V200- V8lx2)
115. (6) Let the two ball meet at t = t sec. 1
Then, first ballget ' sec and second ball get 2 58x36
(t-2) sec yg-2010/2 -9W2) =242
2 45 1
andthey willmeet at same height. x10x 36 ..g=8m/

So, h = h,
Part-B (JEE Advanced)

4. No, Ata given instant of time, the body is

1. |Average velocity |
(b) at two different positions.A and B in the given Also c
| displacement position time graph, which is not possible
time Timc4 ie., they-intercept is negative
Hence graph (a) correctly depics
2r
2 2 m's. (. r=lm;t=Is) corresponding a-x graph.
6. (b) Change in velocity = area under t
2. (2R, nR) Displacement = shortest distance acceleration-time graph
Pos1tion
=AB =AOB=2R
5. (a) The oquation for the given -xgraph is x10x||= 55 m/s
distance
velocityit
(1) Since, initial velocity is zero, final
maximum velocity is 55 m/s.
B dv 7. (8) 2 ms
displacement 0.3 ms 0.2 m s ;
dx
2Tr
Distance- Path length -ACB = =IR 4 m
dx
(b) Average acceleration from (i) For ball A
3.
u, = 0.3 ms
a=
a =: .. (1)
t

On comparing the equation (ii) with equation of X=0.31 -2

a straight line
For ballB
2ms2. s, =4-x,,
W+ s =0.2 ms, a, -
we get m= =+ve,
2
y=+5 ms 4-x= 0.2 +2
i.e. tan =+ ve, 1.e.. 1s acute.
|ATe +=vs+s² -s/Em's Adding eq. (i) and (ii)
4
8s.
la=
|åv| SV2 ms in North-west 4=0.5t
0.5
10 movev easf;
At
Note ::Particle'A' wwill never
locire

initia!
direction. ² is much larger than its
 Line
Motion in a Straight A287

ms. But to match the official answer, we solved (11) Total distance travelled= area ofA ABC 12. (False) For both trains, we have different
as shown above. centrifugal force. So we will have different net
8. (a. c. d) Att=0and =1 body is at rest x base x altitude =-XIX Vmax force acting towards centre.
2
initially a is positive so that the body acquires So, pressure exerted will be different.
some velocity. Then a should be negative so 1 aß 13. (b) For a moving observer, the near by
.Lot the body coOmes to rest. Hence a cannot a+ß 2la+B objects appear to move in the opposite direction
remain positive for all timein the interval 0 < t<1 at a large speed. This is because the angular
10. (5) speed of the near by object w.r.t observer
is
The journey is depicted in the following wt graph. large. As the object moves away the angular
Total time of journey= lsec velocity decreases and therefore its speed
object almost
Txal di_placement = I m= Area under (r-) graph seems to be less. The distant
remains stationary.
=5X'max x1 V
oCOS 30
s i n3 0 °

2s 2x1 = 2 m's
lnax

)3 30 60
p
2 m/s
As 'A'see the motion of"B Lr to V.
So, v =V¡ cos 30° ..0)

=V¡ Cos 301+Vg Sin 30j - v From the concept of relative velocity
For path 00. acceleration (a) [From (1)] V2 =V2G -VG
changein velocity 4 m/s Where Gis the laboratory frame.
time 12 sin 30°j [From (1))
cos 30°
14. (45°, 2m/s) (a) For the ball B to hit the
For path OS is retardation 4m/s?
-Vtan 30° ) 4- 100 ms trolley-A retative velocity of B w.r.t. 4, VgA
For path OP. a (acceleration) >4 ms? should be along O4. So, Vp Willmake an angle
For path /PS (acceleration) 4m/s? S00 45° with +(ve) X-axis.
so, lo Zsec
For path OR, acceleration a < m/s?
4
B4 100 (b) Here 9 =45°
For path RS, retardation a>4 m's' 40 4 x45
Hence a >4 at somc point or points in its path. 60°
11. Each person K, L, M, Nmoves with
9. (i) Let t, be the time taken by the car to
with tve X-axis.
attain the maximum velocity while it is a uniform spced v such that K always move
acceleration.
diretly towards L, Lherectly towards M. M In AOMA,
drectly towards N and N directly towards K. 9-45°04M= 45°
Using v-ut at Here on the bas1s of synmetry we can say that
K, L. M and N will meet at the centre of the
(0) VAG
=0+ al or square O
Since the total time elapsed is . the car At any instant velocity component along KO 135°145._.

decelerators for time t, = (!- I)to come by 454

V
COs 45° =
rest, a=B and v=0
45
Usint v ut al, d
Distance K0= dcos 45° = M

0=-B-I) or =tt ....(u) 204B = 135°

B Also ZB0A - 60°- 45°= 15°
Using (i) in (ii), we get Using sine law in AOBA
or =Vm +

sin|35° sinI5°
taß (0)
or Vm =
(a+B)
VB
V,sinl35° (3-1)x 2v2 2 m/s
velocity distance sin15°
Time taken to meet at '0'
velocity Hence speed of bailw.r.t ground, V¡-2 m/s
d v2 d 15. (True) As the ball falls back to point of
Or, = projection, so s = 0
2-u-2 g*s- 0
, *I-1’ time