# Motion Under Variable Acceleration

## Trending Questions

**Q.**

The displacement x of a particle moving in one dimension is related to time t by the equation t=√x+3 where x is in metres and t in seconds. The displacement of the particle when its velocity is zero is

zero

4 m

1 m

0.5 m

**Q.**

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms^{-1} to 20 ms^{-1} while passing through a distance of 135 m in â€˜tâ€™ seconds. The value of t is:

12

9

10

1.8

**Q.**A particle moves a distance x in time t according to equation x=(t+5)–1. The acceleration of particle is proportional to

- (velocity)3/2
- (distance)2
- (velocity)2/3
- (distance)−2

**Q.**The position x of a particle with respect to time t along x-axis is given by x=9t2−t3, where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x direction?

- 54 m
- 81 m
- 24 m
- 32 m

**Q.**The relation between time and distance is t=αx2+βx, where α and β are constants. The retardation is

- 2βv3
- 2αβv3
- 2β2v3
- 2αv3

**Q.**

A particle has initial velocity $(2\hat{i}+3\hat{j})$ and acceleration $(0.3\hat{i}+0.2\hat{j})$ the magnitude of the velocity after $10s$ will be:

$9\sqrt{2}units$

$5\sqrt{2}units$

$9units$

$5units$

**Q.**

An object moving with a speed of 6.25 m/s, is decelerated at a rate given by:

dvdt=−2.5√v

where v is instantaneous speed. The time taken by the object, to come to rest, would be: ** **

- 2 s
- 4 s
- 1 s
- 8 s

**Q.**

The position $x$ of a particle varies with time as $x=a{t}^{2}-b{t}^{3}$. The acceleration of a particle is zero at time $T$equal to

$\frac{2a}{3b}$

$\frac{a}{b}$

$\frac{a}{3b}$

$0$

**Q.**How acceleration is equal to d^2x / dt^2

**Q.**A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P0. The instantaneous velocity of this car is proportional to

- t−1/2
- t1/2
- t/√m
- t2P0

**Q.**The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v0 The distance travelled by the particle in time t will be

- v0t+13bt2
- v0t+13bt3
- v0t+16bt3
- v0t+12bt2

**Q.**

The velocity of a body depends on time according to equation V=20 + 0.1 t2. the body is undergoing

Uniform Acceleration

Uniform Retardation

Non-Uniform acceleration

Zero Acceleration

**Q.**

The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28.

Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

**Q.**A particle of mass 10−2 kg is moving along the positive x-axis under the influence of a force, F(x)=−k2x2, where k=10−2Nm2. At time t=0, it is at x=1.0 m and its velocity v=0. The speed of the particle when it reaches x=05 m is

**Q.**

The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval $t=0tot=5s$ will be.

**Q.**

The displacement of a particle varies with time t, x=ae−at+beβ t where a, b, α and β are positive constants. The velocity of the particle will?

**Q.**

The acceleration of a particle, starting from rest, varies with time according to the relation a = Kt + C, where K and C are constants of motion. The velocity ‘V’ of the particle after a time ‘t’ will be

- Kt+Ct
- 12Kt2+Ct
- 12(Kt2+Ct)
- Kt2+12Ct

**Q.**The position of a particle moving along the x-axis is expressed as x=at3+bt2+ct+d. The initial acceleration of the particle is

- (a+c)
- 2b
- 6a
- (a+b)

**Q.**

The equation of motion of a particle is given by $s=2{t}^{3}\xe2\u20ac\u201c9{t}^{2}+12t+1$, where s and t are measured in cm and sec. The time when the particle stops momentarily is

$1sec$

$2sec$

$1,2sec$

None of the above

**Q.**The motion of a particle is defined by the position vector →r=A(cost+t sint)^i+A(sint−t cost)^j, where t is expressed in seconds.

The position vector and acceleration vector are perpendicular

- at t=√2 s
- at t=1.5 s
- at t=1 s
- at t=0

**Q.**

A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation v=t2−t where t is in seconds and v in m/s. Find the time interval for which the particle retards.

t = 0.5 s to t = 1 s

t = 0 to t = 0.5 s

t > 1 s

t = 0 to t = 1 s

**Q.**

A particle is moving along the x-axis with its coordinate with the time $t$ given by $x\left(t\right)=-3{t}^{2}+8t+10$meter. Another particle is moving along the y-axis with its coordinate as a function of time given by $y\left(t\right)=5-8{t}^{3}$meter. . At $t=1$ second , the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$ . Then $v\left(m{s}^{-1}\right)$ is

**Q.**

The displacement x of a particle varies with time according to the relation x=ab(1−e−bt). Which of the following statements is correct?

At t=1b, the displacement of the particle is ab.

The velocity and acceleration of the particle at t = 0 are a and b respectively.

The particle will come back to its starting point as t→∞.

None of these

**Q.**A particle moves with an initial velocity v0 and retardation αv, where v is it’s velocity at any time t

- The particle will cover a total distance v0α
- The particle will come to rest after a time 1α
- The particle will continue to move for a very long time
- The velocity of the particle will become v02 after a time 1α

**Q.**A body starts from rest at time t=0, the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be

- 65 m/s
- 55 m/s
- 45 m/s
- 35 m/s

**Q.**A point moves in a straight line under the retardation av2. If the initial velocity is u, the distance covered in 't' seconds is

- 1aln(1+aut)
- aln(aut)
- aut
- 1aln(aut)

**Q.**A particle is moving along x-axis whose instantaneous speed is v2=108−9x2. The acceleration of particle is

- −9x m/s2
- −18x m/s2
- −9x2 m/s2
- None of these

**Q.**A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

- 2100msec2 downwards
- 1400msec2
- 2100msec2 downwards
- 700msec2

**Q.**A particle moves along a straight line and its velocity depends on time as v=4t−t2 m/s. Then for first 5 s

- Average velocity is 253 m/s
- Average speed is 10 m/s
- Average velocity is 53 m/s
- Acceleration is 4 m/s2 at t = 0

**Q.**The position of a particle moving along the x-axis is expressed as x=at3+bt2+ct+d. The initial acceleration of the particle is

- 6a
- 2b
- (a+b)
- (a+c)