Solution

NEET-UG - Physics

Worksheet ch-3 Motion in 1 D

1. (a) difference in velocities
Explanation: difference in velocities
2. (c) 72 kmph
Explanation: let length of the train is L
L 2+L

10
= 110

by simplifying we get L= 0.2 km =200 m

and time 10 sec
200
hence speed is = 10
= 20 m/s= 72km/hr

3. (d) 54 km/h

Explanation:
1 2 1 60t
S1 = at = (at)t = 30t
2 2 2

,S2 = 60 × 8t = 480t, S3 = S1 = 30t

S1 +S2 +S3
vav = = 54km/h
t+8t+t

4. (a) 750 m
Explanation: u = 0, a = 2m/s2 , t = 10 sec
1 2 1
∴ s1 = ut + at = 0+ × 2 × 100
2 2

= 100 m
Velocity after 10 sec,
v = u + at = 0 + 2 × 10 = 20 m/s
∴ s2 = v × 30 = 20 × 30 = 600 m

Final velocity = 0, a = -4 m/s2

∴ 0 = v2 + 2as3

0 = (20)2 - 2 × 4 × s3
400
∴ s3 = = 50m
8

∴ s = s1 + s2 + s3
= 100 + 600 + 50 = 750 m
5. (a) 5 s
Explanation: Let the stone remains in the air for t s.
1
From, S = ut +
2
gt
2

Here, u = 0, Hence, S =
1 2
gt
2

Total distance travelled by the stone in the last second is,

1 1 g
D = St − St−1 = gt
2
− g(t − 1)
2
= (2t − 1) ...(i)
2 2 2

Distance travelled by the stone in first three seconds is,

1 2 9
S3 = × g × 3 = g
2 2

According to given problem, D = S3

g 9
∴
2
(2t − 1) =
2
g or 2r - 1 = 9
t=5s
6. (d) 30 s
Explanation: Relative velocity of overtaking = 40 ms-1 - 30 ms-1 = 10 ms-1. Total distance covered with this
relative velocity during overtaking will be = 100 m + 200 m = 300 m.
300m
Time taken by the first train to pass the second train is: = −1
= 30s
10ms

1/9
 7. (b) 29 seconds
Explanation: Displacement in first eight steps = 5 - 3 = 2 metre.
Since the last five steps covering 5-metre land the drunkard fell
into the pit, the displacement prior to this is (11 - 5) metre = 6 metre.
Time is taken for the first eight steps = 8 sec.
6
Time is taken to cover the first six metres of journey = 2
× 8 = 24sec

Time is taken to cover last 5 metres = 5 sec.

Total time = 24 sec + 5 sec = 29 sec
8. (a)

Explanation:

9. (c) I, II, III

Explanation: When a particle is moving with a uniform velocity, it must move along a straight line in the
same direction. Then its uniform velocity = average velocity and instantaneous speed = instantaneous
velocity. Also, the magnitude of distance is equal to displacement.
−−−−
2
a
10. (b) √ 2 2
v −v
1

Explanation:

a
t = ′
v
a
=
2 2
√v − v
1

−−−−
2
a
= √ 2
2
v −v
1

11. (b) 5 m
a
Explanation: s = u+ (2n − 1)
2

u = 100 m/s, a = -10 m/s2, s=5m

10
5 = 100 − (2n − 1) or n = 10
2

Body thrown upwards with velocity 100 m/s takes 10 sec to reach the highest point.
So body thrown upwards with velocity of200 m/s will take 20 sec to reach the highest point. Hence,
distance travelled in 20th sec,
10
s = 200 − (20 × 2 − 1)
2

= 200 - 5 × 39 = 200 - 195 = 5 m

2/9
 [Note: Bodies travel the same distance in the last second of their upward journey irrespective of their
velocities.]
12. (d) 80 m
Explanation: 400 − s =
1 2
gt ...(i)
2
1
and s = 50t −
2
gt ...(ii)
2

Adding, 50t = 400

or t = 8 sec
1
∴ s = 50 × 8 − × 10 × 64
2

= 400 - 320 = 80 m
13. (a) if the body is slowing, the acceleration is negative
Explanation: If a body is slowing, the acceleration is negative but the distance travelled is not necessarily
negative.
14. (d) 9 s
Explanation: If the relative initial velocity, relative acceleration and relative displacement of the second
body with respect to the first body be ur, ar and sr, then

sr = urt + (1/2)art2
But ur = u2 - u1 = 2 - 0; ∴ ur = 2 m/s
ar = a2 - a1 = 9.8 - 9.8 = 0 and sr - s2 - s1 = 18 m
1
∴ 18 = 2t +
2
(0)t
2
or 18 = 2t or t = 9 sec.
15. (b) 300 m
Explanation: The initial velocity of the ball is = 20 m/s
The final velocity of the ball is = 80 m/s
The acceleration due to gravity is = 10 m/s2
So, the height of the tower can be obtained from the expression:
v2 = u2 + 2gh
Substitute the values in above expression:
802 = 202 + (2 × 10 × h)
6400−400
h= 20
m
6000
h= ⇒ 300 m
20

16. (a) Velocity is zero but speed increases

Explanation: Velocity is zero as total displacement is zero but speed increases as it is being accelerated.
17. (d) 1 : 3 : 5
Explanation: Sn ∝ (2n - 1).
In equal time interval of 2 seconds. Ratio of distance = 1 : 3 : 5
18. (d) 2400 m
a
Explanation: Distance covered in nth second Sn = u+
2
(2n − 1)

where u and a are the initial velocity and acceleration respectively.

Case 1 : Sn = 65, n = 5
a
So, 65 = u+ (2 × 5 − 1)
2

We get 65 = u + 4.5a ...(i)

Case 2 : Sn = 105, n = 9
a
So, 105 = u + 2 (2 × 9 − 1)
We get 105 = u + 8.5a ...(ii)
Equation (ii) -(1), 40 = 4a
Or a = 10 m/s2
So, 65 = u + 4.5 × 10 ⇒ u = 20 m/s
1
Distance covered in 20 seconds, S = ut +
2
2
at

3/9
 where t = 20 s
S = 20 × 20 +
1
(10)(20)
2
= 2400 m
2

19. (a) 0
displacement
Explanation: Average velocity =
total time taken

Displacement = 0, as boat comes back at the starting point.

∴ Average velocity = 0

20. (c) 1260 ms-2

Explanation: v2 = u2 + 2as
−−
−
As u= 0, hence v2 = 2as or v = √2as
When the tennis ball is dropped, the velocity gained by it, just before colliding with the floor is:
−−−− −− −−−
u = √2 × 9.8 × 4 = 8.85 ms-1 (upwards)
= -8.85 ms-1 (downwards)
As the ball rebounds to a height of 2m, hence velocity of the ball just after colliding with the floor is:
−−−− −− −−−
v = + √2 × 9.8 × 2 = 6.26 ms-1 (upwards)
Time of contact of the ball with the floor (given),
t = 12 × 10-3 s
Hence, average acceleration
v−u 6.26−(−8.85)
a = = −3
t
12×10

= +1260 ms-2
21. (b) 550 m
Explanation: u = 0, v = 27.5 m/s, t = 10 sec
v−u 27.5 2
a = = = 2.75m/s
t 10
1 2
s = ut + at
2
1
= 0+
2
× 2.75 × (10 + 10)
2
= 2.75 × 200 = 550 m
3
22. (a) 7
s

Explanation: As ,we know the equation of motion is S = u t + 1/2 at2.....(1)

In case of falling u = 0 and a = g
also S= (50+40)=90 cm =0.90 m
Put these values in equation 1 we get,
9.8 × t2
1
0.90 = 2
×

t = 3/7 sec

23. (d)

1
Explanation: s = ut + at
2

2
1 2
s = at (∵ u = 0)
2

It is an equation of parabola.

24. (c) 15 × 104 m/s2

Explanation: Using the formula, v2 = u2 + 2as
2 2 10
(100) = (200) − 2a ×
100

4/9
 = (200)2 - (100)2 = 300 × 100
10
2a ×
100
5
3×10 4 2
a = = 15 × 10 m/s
2

25. (a) n2s

Explanation: v2 = u2 + 2as or v2 - u2 = 2as
2
v
Maximum retardation, a =
2s
, When the initial Velocity is v, then the distance over which it can be
stopped is given by:
2 2
u (nv)
0 2
sn = = 2
= n s
2a 2(v /2s)

26. (b)

Explanation:

27. (b) in the same direction

Explanation: The magnitude of average velocity is equal to the average speed when total distance
travelled is equal to displacement of particle .This happens when a particle moves in the same direction.
28. (c) (A) only
Explanation: A particle can have only one speed at one time and it cannot have two speeds at one time.
29. (b) equal to or less than one
Explanation: equal to or less than one
30. (a) both the blocks will reach the ground with same speed but the freely falling block first
Explanation: In case of sliding motion on an inclined plane,
1 2
h = s = g sin θ t
2
−−
1 2h −−− −−− −−−− −
or t = tS = √
g
and vS = √2gh = √2(g sin θ)s
sin θ

while in case of free fall,

−−
2h −−−
tF = √
g
and vF = √2gh

tF
∴ = sin θ < 1, i. e, tF < tS
tS

i. e., the falling body reaches the ground first.

v
Also, v = 1, i.e., vF = vS
F

i. e., both reach the ground with the same speed (not velocity as for falling body direction is vertical while
for the sliding body along the plane downwards).
31. (a) 0, 10 m/s
Explanation: Radius of lap = 100m
Circumference = 2πR = 200π = 628m
time taken to cover one lap = 62.8s
628
Hence average speed = 62.8
= 10ms
−1

Average velocity = 0 as the inital and final positions are same (after completing a lap) (displacement = 0)

5/9
32. (a)

Explanation:
Taking initial position as origin and direction of motion(I.e., Vertically up) as positive. As the particle is
thrown with initial Velocity, at the highest point its velocity is zero and then it returns back to its reference
position. This situation is best depicted in figure of option this.
In figure, AB part denotes upward motion and BC part denotes downward motion.

−−−−−
33. (d) √π 2 + 4m

Explanation: Displacement in horizontal direction = nR = n m. Displacement in vertical direction = 2R = 2

m
−−−− −
∴ Resultant displacement = √π 2 + 4 m

34. (a) < 1

Explanation: Since motion is on a curved path, magnitude of displacement is less then the distance
covered.

35. (d) (velocity)3/2

Explanation: Distance, x = (t + 5)-1 ...(i)
dx d
Velocity, v = = (t + 5)
−1

dt dt

= -(t + 5)-2 ...(ii)

dv d
Acceleration a =
dt
=
dt
[−(t + 5)
−2
]

= 2(t + 5)-3 ...(iii)

From equation (ii), we get,
v3/2 = -(t + 5)-3 ...(iv)
Substituting this in equation (iii), we get,
acceleration, a = -2v3/2
or a ∝ (velocity)3/2
From equation (i), we get,
x3 = (t + 5)-3
Substituting this in equation (iii), we get,
Acceleration, a = 2x3
or a ∝ (distance)3 or a ∝ v3/2 [From eqn. (iv)]
36. (c) 45 km/h
d d
Explanation: t =
vre l
=
vc +v
5
i.e., 4
= or vc = 45 km/h
60 vc +30

37. (a) Symmetric parabola

Explanation: Symmetric parabola
38. (d) C
Explanation: Because the slope is the highest at C

6/9
 ds
v = is maximum.
dt

39. (a) 6 s
1
Explanation: S = ut + gt
2

or t2 - 5t - 6 = 0
10 2
30 = −25t + t
2

or (t - 6)(t + 1) = 0 (take positive root)

∴ t = 6 sec

40. (a)

Explanation: In the portion OA, slope (= velocity) of the curve is +ve; at the point A, slope of the curve is
zero; while in the portion AB, slope of the curve is -ve. Hence, (u- t) curve will be as shown in this option.

41. (d) 4.0 m/s

x/2 x
Explanation: t1 = =
3 6

x1 = 4.5t2, x2 = 7.5t2
x
Also, the other half distance is :- x1 + x2 =
2
= (4.5 + 7.5)t2
x
That is t2 =
24
x 2x x
t = t1 + 2t2 = + =
6 24 4
x
v = = 4m/s
t

–
42. (d) 2√2 m
Explanation: Total time of motion on square track = 3 min = 3 × 60 = 180 s
Time period of revolution = 40 s
Therefore, displacement in time = (4 × 40 = 160 sec) = 0, as toy cyclist will be reaching at the starting point.
Thus, displacement in 180 sec. = displacement in 20 sec
−− −−−− –
2 2
= √2 + 2 = 2√2m

43. (d) 16 m
Explanation: x = 40 + 12t - t3
dx
∴ Velocity, v = = 12 − 3t
2

dt

When particle comes to rest,

dx
= v = 0
dt

∴ 12 - 3t2 = 0
or 3t2 = 12 or t = 2 sec
Distance travelled by the particle before coming to rest:
s 2
∫ ds = ∫ vdt
0 0
2
2 3
3t
∴ S = ∫
0
(12 − 3t ) dt = [12t −
2

3
] = 12 × 2 − 8 = 16 m
0

7/9
44. (c) vA = vB = vC

Explanation:

When A is thrown up, it reaches to maximum height at zero velocity, comes back to A with the same initial
velocity vA. vB has the same initial velocity. The vertical velocity for C = 0. vc is acting horizontally.
−−−−−−−
For A: final velocity = √v
2
A
+ 2gh

−−−−−−−
For B: final velocity = √v
2
B
+ 2gh

−− −−−−−
For C: final velocity 2
= √vc + 2gh

As vA = vB = vC (given)
Hence, vf for A = vf for B = vf for C.

45. (c) 32 m
Explanation: 1st body moves with velocity u= 8m/s
1
by using , s= ut + at
2
as a=0
2

then s= 8t .....(1)
for 2nd body :- s= ut +1/2 at2 as u=o and t=4 m/s2
1
then s= × 4× t
2

S= 2 t2 ....(2)
from (1) and (2)nd equation ,we get
t= 4 sec
then displacement s = 4 x 8 = 32 m.
46. (c) 200 m
Explanation: h∝ u2
2
h1 u
1
= 2
h2 u
2
2
u
50 1
= 2
h2 4u
1

⇒ h2 = 200 m
47. (d) 15 m/s
Explanation: Velocity of the thief's car with respect to ground is,
vTG = 10 m/s
Velocity of police man with respect to ground = vPG = 5 m/s
Velocity of bullet fired by police man with respect to ground,
72×5
vBP = 72km/h = = 20m/s
18

Velocity with which bullet will hit the target is,

vBT = vBG + vGT
= vBP + vPG + vGT
= 20 + 5 - 10 = 15 m/s
48. (c) Displacement
Explanation: Displacement
49. (b) 8 s
Explanation: Let the man will be able to catch the bus after t s.
1
then 10t = 48 +
2
× 1× t
2

t2 - 20t + 96 = 0
or, (t - 12)(t - 8) = 0
r = 8s and t = 12s
Thus, the man will be able to catch the bus after 8 s.

8/9
50. (a) 62.4 m
Explanation: h, is the height of balloon by using equation of motion
S= u t + 1/2 a t2
-h = 4 × 4 +1/2 × (-9.8) × 42
-h = 16 -78.4
-h= -62.4
hence, h = 62.4m
51. (b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
Explanation: Both assertion and reason are true but reason is not the correct explanation of assertion.
If a particle travels with speeds v1, v2, v3, .... during time intervals t1, t2, t3, ....respectively.
Then, total distance travelled = v1t1 + v2t2 + v3t3 + ......
total time taken = t1 + t2 + t3 + ...............
v1 t1 +v2 t2 +v3 t3 +…
So, average speed Vav =
t1 +t2 +t3 +…

If, t1 = t2 = t3 = ....... = t
(v1 +v2 +v3 +….)t
Then, Vav. =
nt
v1 +v2 +v3 +…
or Vav, =
n

It means that the average speed is equal to the arithmetic mean of individual speeds.
52. (a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: If both assertion and reason are true and the reason is the correct explanation of assertion.
53. (c) If the assertion is true but the reason is false.
Explanation: The speedometer of an automobile measures the instantaneous speed of the automobile.
54. (a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: If both assertion and reason are true and the reason is the correct explanation of assertion.
55. (c) If the assertion is true but the reason is false.
Explanation: The assertion is true but the reason is false.
In uniform motion, the object moves with uniform velocity, the magnitude of its velocity at different
instants, i. e., at t = 0, t = 1 sec, t = 2 sec,............. , will always be constant. Thus, a velocity-time graph for an
object in uniform motion along a straight path is a straight line parallel to the time axis.
56. (a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: If both assertion and reason are true and the reason is the correct explanation of assertion.
57. (a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: If both assertion and reason are true and the reason is the correct explanation of assertion.
58. (a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: If both assertion and reason are true and the reason is the correct explanation of assertion.
59. (b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
Explanation: As the object is farther away, the angle subtended by the object to an observer in moving
train is small as compared to the angle subtended by nearer objects. As a result, nearer objects appear to
move faster while farther/distant objects seem to be stationary.
The reason is also correct but not explaining Assertion.
60. (b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
Explanation: If both assertion and reason are true but reason is not the correct explanation of assertion.
Since, velocity and acceleration are vector quantities, therefore, the equation of motion is a vector equation
→
(v ⃗ = u⃗ + at ) . This equation is valid when the acceleration direction is the same as velocity.
dv
Acceleration is the rate of velocity, i. e, a = or v = ∫ adt , in order to get the given equation
dt

acceleration should be independent of time, i. e., constant.

9/9