Topic-wise DPPs for NEET Aspirants

Motion in 1-Dimension
DPP- I – Distance, Displacement, Speed, Velocity and Acceleration
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1. A boy walks to his school at a distance of 6𝑘𝑚 with constant speed of 2.5 𝑘𝑚/ℎ𝑜𝑢𝑟 and walks back with a
constant speed of 4 𝑘𝑚/ℎ𝑟. His average speed for round trip expressed in 𝑘𝑚/ℎ𝑜𝑢𝑟, is
a) 24/13 b) 40/13 c) 3 d) 1/2
2. 2
The velocity of a body depends on time according to the equation 𝑣 = 20 + 0.1𝑡 . The body is undergoing
a) Uniform acceleration b) Uniform retardation
c) Non-uniform acceleration d) Zero acceleration
3. The motion of a particle is described by the equation 𝑢 = 𝑎𝑡. The distance travelled by the particle in the
first 4 seconds
a) 4𝑎 b) 12𝑎 c) 6𝑎 d) 8𝑎
4. 1/2
If the velocity of particle is given by 𝑣 = (180 − 16𝑥) 𝑚/𝑠, then its acceleration will be
a) Zero b) 8 𝑚/𝑠 2 c) −8 𝑚/𝑠 2 d) 4 𝑚/𝑠 2
5. The numerical ratio of average velocity to average speed is
a) Always less than one b) Always equal to one
c) Always more than one d) Equal to or less than one
6. 2
If the velocity of a particle is (10 + 2𝑡 )𝑚/𝑠, then the average acceleration of the particle between 2𝑠 and
5𝑠 is
a) 2𝑚/𝑠 2 b) 4𝑚/𝑠 2 c) 12𝑚/𝑠 2 d) 14𝑚/𝑠 2
7. Select the incorrect statements from the following
S1 : Average velocity is path length divided by time interval
S2 : In general, speed is greater than the magnitude of the velocity
S3 : A particle moving in a given direction with a non-zero velocity can have zero speed
S4 : The magnitude of average velocity is the average speed
a) S2 and S3 b) S1 and S4 c) S1, S3 and S4 d) All four statements
8. A bird flies for 4 𝑠 with a velocity of |𝑡 − 2|𝑚/𝑠 in a straight line, where 𝑡 is time in seconds. It covers a
distance of
a) 2 𝑚 b) 4 𝑚 c) 6 𝑚 d) 8 𝑚
9. A particle travels 10𝑚 in first 5 𝑠𝑒𝑐 and 10𝑚 in next 3 𝑠𝑒𝑐. Assuming constant acceleration what is the
distance travelled in next 2 𝑠𝑒𝑐
a) 8.3 𝑚 b) 9.3 𝑚 c) 10.3 𝑚 d) None of above
10. The acceleration ‘𝑎’ in 𝑚/𝑠 2 of a particle is given by 𝑎 = 3𝑡 2 + 2𝑡 + 2 where 𝑡 is the time. If the particle
starts out with a velocity 𝑢 = 2𝑚/𝑠 at 𝑡 = 0, then the velocity at the end of 2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 is
a) 12 𝑚/𝑠 b) 18 𝑚/𝑠 c) 27 𝑚/𝑠 d) 36 𝑚/𝑠
11. The numerical ratio of displacement to the distance covered is always
a) Less than one b) Equal to one
c) Equal to or less than one d) Equal to or greater than one
12. A car moves a distance of 200 𝑚. It covers first half of the distance at speed 60 𝑘𝑚ℎ−1 and the second half
at speed 𝑣. If the average speed is 40 𝑘𝑚ℎ−1 , the value of 𝑣 is
a) 30 𝑘𝑚ℎ−1 b) 13 𝑘𝑚ℎ−1 c) 60 𝑘𝑚ℎ−1 d) 40 𝑘𝑚ℎ−1
13. The displacement of a particle is given by 𝑦 = 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 − 𝑑𝑡 4 . The initial velocity and acceleration are
respectively
a) 𝑏, −4𝑑 b) −𝑏, 2𝑐 c) 𝑏, 2𝑐 d) 2𝑐, −4𝑑
14. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by 𝑑𝑣 = −2.5√𝑣 where 𝑣 is the
𝑑𝑡
instantaneous speed. The time taken by the object, to come to rest, would be
a) 2 s b) 4 s c) 8 s d) 1 s
15. A point initially at rest moves along 𝑥-axis. Its acceleration varies with time as 𝑎 = (6𝑡 + 5)𝑚/𝑠 2 . If it
starts from origin, the distance covered in 2 𝑠 is
a) 20 𝑚 b) 18 𝑚 c) 16 𝑚 d) 25 𝑚
16. The position 𝑥 of a particle with respect to time 𝑡 along 𝑥-axis is given by 𝑥 = 9𝑡 2 − 𝑡 3 where 𝑥 is in
metres and 𝑡 in second. What will be the position of this particle when it achieves maximum speed along
the +𝑥 direction
a) 32 𝑚 b) 54 𝑚 c) 81 𝑚 d) 24 𝑚
17. Acceleration of a particle can be defined when
a) Direction of velocity changes b) Magnitude of velocity changes
c) Both of above d) Speed changes
18. The acceleration of a particle increases linearly with time 𝑡 as 6𝑡. If the initial velocity of the particle is
zero and the particle starts from the origin, then the distance travelled by the particle in time 𝑡 will be
a) 𝑡 b) 𝑡 2 c) 𝑡 3 d) 𝑡 4
19. A body starts from origin and moves along 𝑥-axis such that at any instant velocity is 𝑣𝑡 = 4𝑡 3 − 2𝑡 where 𝑡
is in second and 𝑣𝑡 in ms−1 . The acceleration of the particle when it is 2m from the origin is
a) 28ms−2 b) 22ms −2 c) 12ms −2 d) 10ms −2
20. The position of a particle moving along x-axis at certain times is given below:
𝑡(𝑠) 0 1 2 3
𝑥(𝑚) -2 0 6 16
Which of the following describes the motion correctly
a) Uniform accelerated
b) Uniform decelerated
c) Non-uniform accelerated
d) There is not enough data for generalization
21. The motion of a particle along a straight line is described by equation :
𝑥 = 8 + 12𝑡 − 𝑡 3
Where 𝑥 is in metre and t in second. The retardation of the particle when its velocity becomes zero, is
a) 24𝑚𝑠 −2 b) Zero c) 6𝑚𝑠 −2 d) 12𝑚𝑠 −2
22. A car travels half the distance with constant velocity of 40 𝑘𝑚𝑝ℎ and the remaining half with a constant
velocity of 60 𝑘𝑚𝑝ℎ. The average velocity of the car in 𝑘𝑚𝑝ℎ is
a) 40 b) 45 c) 48 d) 50
23. 2
The relation between time and distance is 𝑡 = 𝛼𝑥 + 𝛽𝑥, where 𝛼 and 𝛽 are constants. The retardation is
a) 2𝛼𝑣 3 b) 2𝛽𝑣 3 c) 2𝛼𝛽𝑣 3 d) 2𝛽 2 𝑣 3
24. The velocity of particle is 𝑣 = 𝑣0 + g𝑡 + 𝑓𝑡 2 . If its position is 𝑥 = 0 at 𝑡 = 0, then its displacement after unit
time (𝑡 = 1) is

a) 𝑣0 + 2g + 3𝑓 b) 𝑣0 + g/2 + 𝑓/3 c) 𝑣0 + g + 𝑓 d) 𝑣0 + g/2 + 𝑓

25. The acceleration ‘𝑎’ in 𝑚/𝑠 of a particle is given by 𝑎 = 3𝑡 + 2𝑡 + 2 where 𝑡 is the time. If the particle
2 2

starts out with a velocity 𝑢 = 2𝑚/𝑠 at 𝑡 = 0, then the velocity at the end of 2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 is
a) 12 𝑚/𝑠 b) 18 𝑚/𝑠 c) 27 𝑚/𝑠 d) 36 𝑚/𝑠
26. A particle is moving on a straight line path with constant acceleration directed along the direction of
instantaneous velocity. Which of the following statements are false about the motion of particle?
a) Particle may reverse the direction of motion
b) Distance covered is not equal to magnitude of displacement
c) The magnitude of average velocity is less than average speed
d) All of the above
27. Which of the following 4 statements is false
a) A body can have zero velocity and still be accelerated
b) A body can have a constant velocity and still have a varying speed
c) A body can have a constant speed and still have a varying velocity
d) The direction of the velocity of a body can change when its acceleration is constant
28. A car travels equal distances in the same direction with velocities 60kmh−1 , 20 km h−1 and 10
km h−1 respectively. The average velocity of the car over the whole journey of motion is
a) 8 ms−1 b) 7 ms−1 c) 6 ms−1 d) 5 ms−1
 : HINTS AND SOLUTIONS :
1 (b) 𝐯av displacement
2𝑣1 𝑣2 2×2.5×4
= ≤1
Distance average speed = = 𝑣av length of path (distance)
𝑣1 +𝑣2 2.5+4
200 40 6 (d)
= = 𝑘𝑚/ℎ𝑟
65 13 Change in velocity 𝑣 −𝑣
Average acceleration = = 𝑡2 −𝑡 1
2 (c) Time taken 2 1
𝑑𝑣 [10 + 2(5)2 ] − [10 + 2(2)2 ] 60 − 18
Acceleration = 𝑎 = = 0.1 × 2𝑡 = 0.2𝑡 = = 14𝑚/𝑠 2
𝑑𝑡 3 3
Which is time dependent 𝑖. 𝑒. non-uniform 7 (c)
acceleration Displacement
Average velocity = Time interval
3 (d)
A particle moving in a given direction with non-
𝑎𝑡 2
𝑢 = 𝑎𝑡, 𝑥 = ∫ 𝑢 𝑑𝑡 = ∫ 𝑎𝑡 𝑑𝑡 = zero velocity cannot have zero speed.
2 In general, average speed is not equal to
For 𝑡 = 4 sec, 𝑥 = 8𝑎
magnitude of average velocity. However, it can be
4 (c)
so if the motion is along a straight line without
𝑣 = (180 − 16𝑥)1/2 change in direction
𝑑𝑣 𝑑𝑣 𝑑𝑥
As 𝑎 = = 𝑑𝑥 . 𝑑𝑡 8 (b)
𝑑𝑡
1 𝑑𝑥 The velocity time graph for given problem is
∴𝑎= (180 − 16𝑥) −1/2 × (−16) ( )
2 𝑑𝑡 shown in the figure.
= −8(180 − 16𝑥)−1/2 × 𝑣
= −8(180 − 16𝑥)−1/2 × (180 − 16𝑥)1/2
= −8 𝑚/𝑠 2
5 (d)
The average speed

length of path 𝐴𝐶𝐵

𝑣av = … (i)
time interval (𝑡2 − 𝑡1 ) Distance travelled 𝑆 = Area under curve = 2 +
2 = 4𝑚
9 (a)
Let initial (𝑡 = 0) velocity of particle= 𝑢
For first 5 𝑠𝑒𝑐 of motion 𝑠5 = 10 𝑚𝑒𝑡𝑟𝑒
1 1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2 ⇒ 10 = 5𝑢 + 𝑎(5)2
2 2
2𝑢 + 5𝑎 = 4
For first 8 sec of motion 𝑠8 = 20 𝑚𝑒𝑡𝑟𝑒
And average velocity, 1
20 = 8𝑢 + 𝑎(8)2 ⇒ 2𝑢 + 8𝑎 = 5
2
displacement 𝐫2 − 𝐫1 7 1
𝐯av = = … (ii) By solving 𝑢 = 6 𝑚/𝑠 and 𝑎 = 3 𝑚/𝑠 2
time interval 𝑡2 − 𝑡1
Now distance travelled by particle in Total 10 𝑠𝑒𝑐
But we know that distance is always be greater 1
𝑠10 = 𝑢 × 10 + 𝑎(10)2
than or equal to magnitude of displacement. So 2
the average speed will always be greater than or By substituting the value of 𝑢 and 𝑎 we will get
equal to the magnitude of average velocity. 𝑠10 = 28.3 𝑚
so the distance in last 2 𝑠𝑒𝑐 = 𝑠10 − 𝑠8
From Eqs. (i) and (ii) = 28.3 − 20 = 8.3𝑚
10 (b) 𝑣 𝑡
Integrating it, we have ∫0 𝑑𝑣 = ∫0 (6𝑡 + 5)𝑑𝑡
𝑣 = 𝑢 + ∫ 𝑎𝑑𝑡 = 𝑢 + ∫(3𝑡 2 + 2𝑡 + 2)𝑑𝑡 𝑣 = 3𝑡 2 + 5𝑡 + 𝐶, where 𝐶 is a constant of
integration
3𝑡 3 2𝑡 2
=𝑢+ + + 2𝑡 = 𝑢 + 𝑡 3 + 𝑡 2 + 2𝑡 When 𝑡 = 0, 𝑣 = 0 so 𝐶 = 0
3 2 𝑑𝑠
= 2 + 8 + 4 + 4 = 18 𝑚/𝑠 (As 𝑡 = 2 𝑠𝑒𝑐) ∴ 𝑣 = 𝑑𝑡 = 3𝑡 2 + 5𝑡 or, 𝑑𝑠 = (3𝑡 2 + 5𝑡)𝑑𝑡
11 (c) Integrating it within the conditions of motion, 𝑖. 𝑒.,
Since displacement is always less than or equal to as 𝑡 changes from 0 to 2𝑠, 𝑠 changes from 0 to 𝑠,
distance, but never greater than distance. Hence we have
numerical ratio of displacement to the distance 𝑠 2
∫ 𝑑𝑠 = ∫ (3𝑡 2 + 5𝑡)𝑑𝑡
covered is always equal to or less than one 0 0
12 (a) 5 2
[𝑡 3 + 𝑡 2 ] = 8 + 10 = 18 𝑚
Time taken by the car to cover first half of the ∴ 𝑠 =
2 0
distance is 16 (b)
100 𝑑𝑥
𝑡1 = 𝑥 = 9𝑡 2 − 𝑡 3 ; 𝑣 = = 18𝑡 − 3𝑡 2 , For maximum
60 𝑑𝑡
Time taken by the car to cover speed half of the speed
distance is 𝑑𝑣 𝑑
= [18𝑡 − 3𝑡 2 ] = 0 ⇒ 18 − 6𝑡 = 0 ∴ 𝑡
100 𝑑𝑡 𝑑𝑡
𝑡2 = = 3 sec
𝑣
Average speed , 𝑣𝑎𝑣 =
Total distance travelled i.e., Particle achieve maximum speed at 𝑡 = 3 sec.
TOtal time taken
At this instant position of this particle, 𝑥 = 9𝑡 2 −
100 + 100 200
𝑣𝑎𝑣 = ⇒ 40 = 100 100 𝑡3
𝑡1 + 𝑡2 + 𝑣
60 = 9(3)2 − (3)3 = 81 − 27 = 54 𝑚
1 1 1 1 1 1 17 (c)
+ = ⇒ = −
60 𝑣 20 𝑣 20 60 Because acceleration is a vector quantity
1 2 1
= = 18 (c)
𝑣 60 30 𝑑𝑣 6𝑡 2
𝑣 = 30𝑘𝑚 ℎ−1 = 6𝑡 or 𝑑𝑣 = 6𝑡, 𝑚𝑣 = = 3𝑡 2 ,
𝑑𝑡 2
13 (c) 𝑡3
2 4 𝑑𝑥 = 3𝑡 2 𝑑𝑡 ⇒ 𝑥 = 3 = 𝑡 3
𝑦 = 𝑎 + 𝑏𝑡 + 𝑐𝑡 − 𝑑𝑡 2
𝑑𝑦 3
∴ 𝑣 = 𝑑𝑡 = 𝑏 + 2𝑐𝑡 − 4𝑑𝑡 and 𝑎 = 𝑑𝑡 = 2𝑐 −
𝑑𝑣 19 (b)
𝑣𝑡 = 4𝑡 3 − 2𝑡
12𝑑𝑡 2
𝑑𝑥𝑡
Hence, at 𝑡 = 0, 𝑣initial = 𝑏 and 𝑎initial = 2𝑐 ⇒ = 4𝑡 3 − 2𝑡
𝑑𝑡
14 (a)
3
𝑑𝑣 ⇒ ∫ 𝑑𝑥𝑡 = ∫ 4𝑡 𝑑𝑡 − ∫ 2𝑡𝑑𝑡
= −2.5 √𝑣
𝑑𝑡
⇒ 𝑥𝑡 = 𝑡 4 − 𝑡 2
𝑑𝑣 Since, 𝑥𝑡 = 2m
⇒ = −2.5𝑑𝑡
√𝑣 𝑡 = √2s (rejecting negative time)
0 𝑡
Now acceleration,
⇒ ∫ 𝑣 −1/2 𝑑𝑣 = −2.5 ∫ 𝑑𝑡 𝑑𝑣𝑡
𝑎𝑡 = = 12𝑡 2 − 2 = 12(2) − 2 = 22ms−2
6.25 0 𝑑𝑡
0
20 (c)
⇒ −2.5[𝑡]𝑡0 = [2𝑣1/2 ]6.25 ∆𝑥
Instantaeneous velocity = 𝑣 =
∆𝑡
⇒ 𝑡 =2s By using the data from the table
0 − (−2) 6−0
15 (b) 𝑣1 = = 2𝑚/𝑠, 𝑣2 = = 6𝑚/𝑠
1 1
Given acceleration 𝑎 = 6𝑡 + 5 16 − 6
𝑑𝑣 𝑣3 = = 10𝑚/𝑠
∴𝑎= = 6𝑡 + 5, 𝑑𝑣 = (6𝑡 + 5)𝑑𝑡 1
𝑑𝑡 So, motion is non-uniform but accelerated

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21 (d) 24 (b)
𝑥 = 8 + 12𝑡 + 𝑡 3 𝑣 = 𝑣0 + g𝑡 + 𝑓𝑡 2
𝑣 = 0 + 12 − 3𝑡 2 = 0
𝑑𝑥
3𝑡 2 = 12 Or = 𝑣0 + g𝑡 + 𝑓𝑡 2
𝑑𝑡
𝑡 = 2 𝑠𝑒𝑐
𝑑𝑣 Or 𝑑𝑥 = (𝑣0 + g𝑡 + 𝑓𝑡 2 ) 𝑑𝑡
𝑎= = 0 − 6𝑡
𝑑𝑡 𝑥 1
𝑎[𝑡 = 2] = −12 𝑚/𝑠 2 So, ∫0 𝑑𝑥 = ∫0 (𝑣0 + g𝑡 + 𝑓𝑡 2 ) 𝑑𝑡
Retardation = 12 𝑚/𝑠 2 g 𝑓
22 (c) Or 𝑥 = 𝑣0 + 2 + 3
2𝑣1 𝑣2 2 × 40 × 60
𝑣𝑎𝑣 = = = 48 𝑘𝑚𝑝ℎ 25 (b)
𝑣1 + 𝑣2 100
23 (a) 𝑣 = 𝑢 + ∫ 𝑎𝑑𝑡 = 𝑢 + ∫(3𝑡 2 + 2𝑡 + 2)𝑑𝑡
𝑑𝑡 1
= 2𝛼𝑥 + 𝛽 ⇒ 𝑣 = 3𝑡 3 2𝑡 2
𝑑𝑥 2𝛼𝑥 + 𝛽 =𝑢+ + + 2𝑡 = 𝑢 + 𝑡 3 + 𝑡 2 + 2𝑡
𝑑𝑣 𝑑𝑣 𝑑𝑥 3 2
∵𝑎= = . = 2 + 8 + 4 + 4 = 18 𝑚/𝑠 (As 𝑡 = 2 𝑠𝑒𝑐)
𝑑𝑡 𝑑𝑥 𝑑𝑡
𝑑𝑣 −𝑣. 2𝛼 26 (d)
𝑎=𝑣 = = −2𝛼. 𝑣. 𝑣 2 = −2𝛼𝑣 3 Since, acceleration is in the direction of
𝑑𝑥 (2𝛼𝑥 + 𝛽)2
∴ Retardation = 2𝛼𝑣 3 instantaneous velocity, so particle always moves
in forward direction.
Hence, (d) is correct.
27 (b)
Constant velocity means constant speed as well as
same direction throughout
28 (d)
3𝑥 3𝑥
Average velocity= 𝑥 𝑥 𝑥 = 𝑥+3𝑥+6𝑥
+ +
60 20 10 60
3𝑥 × 60
= = 18kmh−1
10𝑥
18 × 5 −1
= ms = 5ms −1
18