byju’s home learning program...16. the acceleration- time graph of a particle moving along a...

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Motion in 1D (Session-2) Page | 1

Topic covered:

• Motion in 1-D (Session - 2) - NEET

Worksheet

1. The displacement of a particle varies with time (t) as: 𝑠 = 𝑎𝑡2 − 𝑏𝑡3. Determine the time(𝑡) at which acceleration of the particle becomes 0.

a. 𝑎

3𝑏 b.

𝑎

𝑏

c. 3𝑏

𝑎 d.

2𝑎

3𝑏

2. The displacement travelled by a particle in a straight-line motion is directly

proportional to 𝑡1/2 , where t = time elapsed. What is the nature of motion? a. Increasing acceleration b. Decreasing acceleration c. Increasing retardation d. Decreasing retardation

3. If a particle moves with a constant velocity

a. Its acceleration is positive b. Its acceleration is negative c. Its acceleration is zero d. Its speed is zero

4. The x and y coordinates of a particle at any time t are given by 𝑥 = 7𝑡 + 4𝑡2 and 𝑦 = 5𝑡, where x and y are in 𝑚 and 𝑡 in 𝑠. The acceleration of the particle at 5 𝑠 is. a. Zero b. 8 𝑚/𝑠2 c. 20 𝑚/𝑠2 d. 40 𝑚/𝑠2

5. The velocity displacement curve for an object moving along a straight line is shown in the given figure. At which of the marked point, the object is speeding up?

a. 1 b. 2 c. 1 and 3 d. 1, 2 and 3

6. The velocity – time graph of a particle starting from rest with uniform acceleration is a straight line a. not passing through origin b. parallel to time axis c. parallel to velocity axis d. having none of the above characteristics.


Motion in 1-D (Session-2) Page | 2

7. Given: v = 0.6- 0.3 t (in SI units). The acceleration is a. −0.3 𝑚𝑠−2 b. 0.6 𝑚𝑠−2 c. 2.4 𝑚𝑠−2 d. 3.6 𝑚𝑠−2

8. Given below are the equations of motion of four particles 𝐴, 𝐵, 𝐶 and 𝐷. 𝑥𝐴 = 6𝑡 − 3; 𝑥𝐵 = 4𝑡2 − 2𝑡 + 3; 𝑥𝐶 = 3𝑡3 − 2𝑡2 + 𝑡 − 7; 𝑥𝐷 = 7 cos 60° − 3 sin 30° Which of these four particles move with uniform acceleration? a. A b. B c. C d. D

9. The adjoining figures gives the velocity-time graph. This shows that the body is

a. Starting from rest and moving with uniform velocity b. Moving with uniform retardation c. Moving with uniform acceleration d. Having same initial and final velocity

10. When acceleration be function of velocity as a = f(v), then

a. The displacement (𝑥) = ∫𝑣𝑑𝑣

𝑓(𝑣)

b. The acceleration may be constant c. The slope of acceleration versus velocity graph may be constant. d. (a) and (c) are correct.

11. A particle starts from rest at 𝑡 = 0 and moves in a straight line with acceleration as shown in the following figure.

The velocity of the particle at 𝑡 = 3 seconds is a. 2 𝑚/𝑠 b. 4 𝑚/𝑠 c. 6 𝑚/𝑠 d. 8 𝑚/𝑠



12. Figure (i) and (ii) below show the displacement time graphs of two particles moving along the 𝑥 – 𝑎𝑥𝑖𝑠. We can say that

a. Both the particles are having a uniformly acceleration motion b. Both the particles are having a uniformly retarded motion c. Particle (i) is having a uniformly accelerated motion while particle (ii) is having a

uniformly retarded motion. d. Particle (i) is having a uniformly retarded motion while particle (ii) is having a

uniformly accelerated motion.

13. The position x of a body as a function of time t is given by the equation: 𝑥 = 2𝑡3 − 6𝑡2 + 12𝑡 + 6

The acceleration of the body is zero at time t is equal to a. 1 s b. 2 s c. 3 s d. 0.5 s

14. From the velocity- time graph, given in figure of a particle moving in a straight line, one can conclude that

a. Its average velocity during the 12 𝑠 interval is

24

7𝑚𝑠−1

b. Its velocity for the first 3 s is uniform and is equal to 4 𝑚𝑠−1 c. The body has a non-zero constant acceleration between 𝑡 = 3 𝑠 and 𝑡 = 8 𝑠 d. The body has a uniform retardation from 𝑡 = 8 𝑠 to 𝑡 = 12 𝑠

15. The velocity- time graph of a body is given in Fig. The maximum acceleration in 𝑚𝑠−2 is



a. 4 b. 3 c. 2 d. 1

16. The acceleration- time graph of a particle moving along a straight line is as shown in

figure At what time the particle acquires its initial speed?

a. 12 s b. 5 s c. 8 s d. 16 s

17. For velocity –time graph of given Figure

The acceleration – time graph of the motion of the body is

a. b.

c. d.



18. The displacement (𝑠) of a particle is proportional to the first power of time t, i.e., 𝑠 ∝ 𝑡; then the acceleration of the particle is a. Infinite b. Zero c. A small finite value d. A large finite value

19. The velocity- time graph of a body is shown in the given figure. The body has maximum acceleration during the interval.

a. OA b. BC c. CD d. DE

20. Each of the four particles move along x axis. Their coordinate (in meters) as function of

time (in seconds) are given by Particle 1: 𝑥(𝑡) = 3.5 − 2.7 𝑡3 Particle 2: 𝑥(𝑡) = 3.5 + 2.7 𝑡3 Particle 3: 𝑥(𝑡) = 3.5 + 2.7 𝑡2 Particle 4: 𝑥(𝑡) = 3.5 − 3.4𝑡 − 2.7 𝑡2 Which of these particles are speeding up for 𝑡 > 0 ? a. All four b. Only 1 c. Only 2 and 3 d. Only 2, 3 and 4



Answers Keys

Question

Number 1 2 3 4 5

Answer

Key (a) (d) (c) (b) (a)

Question

Number 6 7 8 9 10

Answer

Key (d) (a) (b) (b) (d)

Question

Number 11 12 13 14 15

Answer

Key (b) (c) (a) (d) (a)

Question

Number 16 17 18 19 20

Answer

Key (c) (c) (b) (b) (a)



Solutions 1. (a) Given,

𝑠 = at2 − bt3

Differentiating 𝑠 w.r.t t we get,

𝑣 =𝑑𝑠

𝑑𝑡= 2𝑎𝑡 − 3𝑏𝑡2

Differentiating 𝑣 w.r.t t we get

𝑎 =𝑑𝑣

𝑑𝑡= 2𝑎 − 6𝑏𝑡

As 𝑎 = 0, we get

2𝑎 − 6𝑏𝑡 = 0 ⟹ 𝑡 =𝑎

3𝑏

2. (d)

𝑆 ∝ 𝑡1

2

⟹ 𝑆 = 𝑘𝑡1

2 (where 𝑘 is a constant)

To get acceleration we need to differentiate 𝑠 twice

⟹ 𝑎 =𝑑2𝑆

𝑑𝑡2 = −1

4𝑘𝑡−3/2

Negative sign indicates that it is retardation and as its value is decreasing with time so, it is decreasing retardation.

3. (c)

Acceleration is the rate of change of velocity

𝑎 =𝑑𝑣

𝑑𝑡

If the particle has to move with constant velocity then

⟹ 𝑑𝑣 = 0

∴ 𝑎 = 0

Acceleration must be 0

4. (b)

To get acceleration along 𝑥 −axis and 𝑦 −axis we need to differentiate both 𝑥 (𝑡) and 𝑦 (𝑡) twice w.r.t 𝑡. 𝑥(𝑡) = 7𝑡 + 4𝑡2

⟹ 𝑎𝑥 =𝑑2𝑥

𝑑𝑡2= 8

and 𝑦(𝑡) = 5𝑡

⟹ 𝑎𝑦 =𝑑2𝑦

𝑑𝑡2 = 0

Hence, net acceleration = √𝑎𝑥2 + 𝑎𝑦

2 = 8 𝑚/𝑠2



5. (a) We know that if acceleration and velocity are in same direction then the speed of the body increases.

Acceleration of the body is given by

𝑎 = 𝑣𝑑𝑣

𝑑𝑠

We can find the sign of acceleration at various points, 𝑣 is +𝑣𝑒, for all three points 1, 2 and 3. 𝑑𝑣

𝑑𝑠 is +ve for point 1, zero for points 2 and – 𝑣𝑒 for point 3.

So, only for point 1, velocity and acceleration have same sign. So, the object is speeding up at point 1 only.

Hence, the correct answer is option (a).

6. (d)

(a)Since initial velocity is zero, therefore, the velocity- time graph must pass through the origin.

(b)Since velocity is increasing at a uniform rate therefore, the velocity- time graph cannot be parallel to time – axis.

(c)If the velocity- time graph is parallel to velocity axis, then it would mean different velocities at any given time. This is physically impossible.

7. (a)

𝑎 = 𝑑𝑣/ 𝑑𝑡 (where 𝑣 is velocity)

Acceleration =𝑑

𝑑𝑡(0.6 − 0.3𝑡)

⟹ 𝑎 = −0.3 𝑚𝑠−2 8. (b)

𝑥𝐴 = 6𝑡 − 3

𝑣𝐴 =𝑑

𝑑𝑡(6𝑡 − 3) = 6

Particle A moves with constant velocity

𝑣𝐵 =𝑑

𝑑𝑡(4𝑡2 − 2𝑡 + 3) = 8𝑡 − 2, 𝑎𝐵 =

𝑑

𝑑𝑡(8𝑡 − 2) = 8

Particle B moves with constant acceleration

𝑣𝐶 =𝑑

𝑑𝑡(3𝑡3 − 2𝑡2 + 𝑡 − 7) = 9𝑡2 − 4𝑡 + 1, 𝑎𝐶 =

𝑑

𝑑𝑡(9𝑡2 − 4𝑡 + 1) = 18𝑡 − 4

So, the particle moves with variable acceleration.

𝑥𝐷 = 7 cos 60° − 3 sin 30°

Since 𝑥𝐷 does not depend upon time therefore particle is at rest.

Thus, only particle A moves with uniform acceleration.

9. (b)

Slope of 𝑣 − 𝑡 graph represents acceleration.

There occurs a decrease of velocity w.r.t time, i.e., body undergoes retardation

Hence, the correct answer is option (b).



10. (d) It is given that 𝑎 = 𝑓(𝑣)

⇒ 𝑣𝑑𝑣

𝑑𝑥= 𝑓(𝑣)

⇒ ∫ 𝑑𝑥 = ∫𝑣𝑑𝑣

𝑓(𝑣)

𝑥

0

⇒ 𝑥 = ∫𝑣𝑑𝑣

𝑓(𝑣)

If acceleration is a linear function of velocity, the 𝑎 − 𝑣 graph will be a straight line, e.g., if 𝑎 = 4𝑣 + 5, then 𝑎 − 𝑣 graph will be straight line. Hence, in this case slope of 𝑎 − 𝑣 graph will be constant. As acceleration is dependent on velocity and acceleration will change 𝑣, so acceleration will also change and can’t remain constant. So, the correct answer is option (d).

11. (b) As object start from rest, area of 𝑎 − 𝑡 curve give change in velocity Velocity at 3 seconds = total algebraic sum of area under the curve. or 𝑣 = 4 × 2 − 4 × 1 = 8 − 4 = 4 𝑚/𝑠 Hence, the correct answer is option (b).

12. (c)

The slope of displacement – time graph gives the velocity. In figure (i), the slope is low in the beginning but it keeps on increasing with time. Thus, velocity of the particle increases with time and hence the particle (i) has accelerated motion. The displacement –time graph has concave shape for accelerated motion.

In Figure (ii), the slope is high in the beginning and it keeps on decreasing with time. Thus, velocity of the particle decreases with time and hence the particle (ii) has retarded motion. The displacement – time graph has convex shape for retarded motion.

For uniformly accelerated or retarded motion, the shape of displacement –time graph is parabolic (which look to be true for given graphs).

13. (a)

𝑥 = 2𝑡3 − 6𝑡2 + 12𝑡 + 6

𝑣 =𝑑𝑥

𝑑𝑡

𝑣 = 6𝑡2 − 12𝑡 + 12

Acceleration =𝑑

𝑑𝑡(𝑣) = 12𝑡 − 12

0 = 12𝑡 − 12

𝑡 = 1 𝑠

14. (d) Displacement in 12 s = area under v – t graph

=1

2× (12 + 5) × 4 = 34 𝑚

𝑉𝑎𝑣 =𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑇𝑖𝑚𝑒=

34

12=

17

6𝑚/𝑠



Hence (a) is incorrect. Option b is incorrect because during first 3s velocity increases from 0 to 4 m/s Option c is incorrect, because in part AB, velocity is constant due to which acceleration becomes 0. From 𝑡 = 8 𝑠 to 𝑡 = 12 𝑠, slope is negative due to which the body is having uniform retardation.

15. (a)

Acceleration from 0 to 20 s will be 𝑎 = 𝑣2−𝑣1

𝑡2−𝑡1=

20−0

20−0= 1


𝑡2−𝑡1=

20−20

30−20= 0


𝑡2−𝑡1=

0−60

70−40= −2

Maximum acceleration will be from 30 to 40 s, because slope in this interval is maximum

𝑎 = 𝑣2 − 𝑣1

𝑡2 − 𝑡1

= 60−20

40−30 = 4 𝑚/𝑠2

16. (c) We know that under area under 𝑎 − 𝑡 graph represents change in velocity.

Let 𝐴1 and 𝐴2 be the areas of upper and lower triangle respectively.

Particle will acquire the initial speed again when,

𝐴1 + 𝐴2 = 0

⟹1

2× 10 × 4 +

1

2× (𝑡0 − 4) × −10 = 0

⟹ 𝑡 = 8 𝑠

At this time direction of velocity will be opposite but the magnitude will be equal.

17. (c) For 0 to 3 𝑠, acceleration is positive as slope of 𝑣 − 𝑡 graph is positive

For 3 to 5 𝑠, acceleration is zero as slope of 𝑣 − 𝑡 graph is 0

From 5 to 7 𝑠, acceleration is negative as slope of 𝑣 − 𝑡 graph is negative.

Hence, option c is correct.

18. (b) Let 𝑆 = 𝑘𝑡 + 𝑐 Differentiating 𝑠 twice to get acceleration, we see that

𝑑2𝑆

𝑑𝑡2 = 0

∴ Acceleration comes out to be zero.



19. (b) The acceleration of the body is the slope of the velocity time graph. This slope is maximum when the graph is steep upwards. That portion in the figure is part BC. Hence, the correct answer is option (b).

20. (a) At t >0 For particles 2 and 3 𝑑𝑥

𝑑𝑡> 0

Also, |𝑑2𝑥

𝑑𝑡2| > 0

As, both velocity and acceleration are positive so particle will be speeding up For particle 1 At 𝑡 > 0

𝑑𝑥

𝑑𝑡= −8.1𝑡2

⟹𝑑𝑥

𝑑𝑡< 0

and 𝑑2𝑥

𝑑𝑡2 = −16.2 𝑡

∴𝑑2𝑥

𝑑𝑡2< 0

For particle 4 At 𝑡 > 0

𝑑𝑥

𝑑𝑡= −3.4 − 5.4𝑡

⟹𝑑𝑥

𝑑𝑡< 0

and 𝑑2𝑥

𝑑𝑡2 = −5.4

∴𝑑2𝑥

𝑑𝑡2 < 0

As, both velocity and acceleration are negative so in this case also body will speed up

Therefore for t > 0; |𝑑𝑥

𝑑𝑡| is increasing in all.

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