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Motion in a Straight Line NEET Questions — Free MCQ Practice

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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

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Explanation

Let initial $(t = 0)$ velocity of particle = u

For first 5 sec motion $s_{5} = 10 m e t r e$

$s = u t + \frac{1}{2} a t^{2} \Rightarrow 10 = 5 u + \frac{1}{2} a {(5)}^{2}$

$2 u + 5 a = 4$ …(i)

For first 8 sec of motion $s_{8} = 20 m e t r e$

$20 = 8 u + \frac{1}{2} a {(8)}^{2} \Rightarrow 2 u + 8 a = 5$ …(ii)

By solving $u = \frac{7}{6} m / s and a = \frac{1}{3} m / s^{2}$

Now distance travelled by particle in Total 10 sec.

$s_{10} = u \times 10 + \frac{1}{2} a {(10)}^{2}$

By substituting the value of u and a we will get $s_{10} = 28 .3 m$

so the distance in last $2 \sec = s_{10} - s_{8}$

$= 28 .3 - 20 = 8 .3 m$

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The distance travelled by a particle is proportional to the squares of time, then the particle travels with

A.

Uniform acceleration

B.

Uniform velocity

C.

Increasing acceleration

D.

Decreasing velocity

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Explanation

$s \propto t^{2}$ (given) ∴ $s = K t^{2}$

Acceleration $a = \frac{d^{s} s}{d t^{2}} = 2 k$ (constant)

It means the particle travels with uniform acceleration.

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Velocity of a particle changes when

A.

Direction of velocity changes

B.

Magnitude of velocity changes

C.

Both of above

D.

None of the above

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Explanation

Because velocity is a vector quantity

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The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 seconds

A.

4 a

B.

12 a

C.

6 a

D.

8 a

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Explanation

$u = a t, x = \int_{}^{} u d t = \int_{}^{} a t d t = \frac{a t^{2}}{2}$

For $t = 4 \sec, x = 8 a$

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The relation $3 t = \sqrt{3 x} + 6$ describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is

A.

24 metres

B.

12 metres

C.

5 metres

D.

Zero

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Explanation

$3 t = \sqrt{3 x} + 6 \Rightarrow 3 x = {(3 t - 6)}^{2}$

$\Rightarrow x = 3 t^{2} - 12 t + 12$

$v = \frac{d x}{d t} = 6 t - 12$ , for $v = 0, t = 2 \sec$

$x = 3 {(2)}^{2} - 12 \times 2 + 12 = 0$

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The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms^–1. If the change in velocity of the body is 0.18ms^–1 during this time, its uniform acceleration is

A.

0.01 ms^–2

B.

0.02 ms^–2

C.

0.03 ms^–2

D.

0.04 ms^–2

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Explanation

$Time = \frac{Distance}{Average velocity} = \frac{3 .06}{0 .34} = 9 \sec$

Acceleration $= \frac{Change in velocity}{Time} = \frac{0 .18}{9}$ =0.02 m/s²

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Equation of displacement for any particle is $s = 3 t^{3} + 7 t^{2} + 14 t + 8 m$ . Its acceleration at time t = 1 sec is

A.

10 m/s²

B.

16 m/s²

C.

25 m/s²

D.

32 m/s²

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Explanation

$s = 3 t^{3} + 7 t^{2} + 14 t + 8 m$

$a = \frac{d^{2} s}{d t^{2}} = 18 t + 14$ at $t = 1 \sec \Rightarrow a = 32 m / s^{2}$

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The position of a particle moving along the 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

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Explanation

Average velocity $v = \frac{Δ x}{Δ t}$

By using the data from the table

$v_{1} = \frac{0 - (- 2)}{1} = 2 m / s . v_{2} = \frac{6 - 0}{1} = 6 m / s$

$v_{3} = \frac{16 - 6}{1} = 10 m / s$

So, motion is non-uniform but accelerated.

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Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process ?

A.

Velocity only

B.

Displacement and velocity

C.

Acceleration, velocity and displacement

D.

Displacement and acceleration

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Explanation

Only direction of displacement and velocity gets changed, acceleration is always directed vertically downward.

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The displacement of a particle, moving in a straight line, is given by $s = 2 t^{2} + 2 t + 4$ where s is in metres and t in seconds. The acceleration of the particle is

A.

2 m/s²

B.

4 m/s²

C.

6 m/s²

D.

8 m/s²

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Explanation

$s = 2 t^{2} + 2 t + 4, v = \frac{d s}{d t} = 4 t + 2 a = \frac{d v}{d t} = 4 m / s^{2}$

Motion in a Straight Line MCQs for NEET — Physics Questions with Answers

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