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

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Which of the following four statements is false ?

A body can have zero velocity and still be accelerated

A body can have a constant velocity and still have a varying speed

A body can have a constant speed and still have a varying velocity

The direction of the velocity of a body can change when its acceleration is constant

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Explanation

Constant velocity means constant speed as well as same direction throughout.

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A particle moving with a uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4 sec each. Its initial velocity is

1 m/sec

10 m/sec

5 m/sec

2 m/sec

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Explanation

Distance travelled in 4 sec

$24 = 4 u + \frac{1}{2} a \times 16$ …(i)

Distance travelled in total 8 sec

$88 = 8 u + \frac{1}{2} a \times 64$ …(ii)

After solving (i) and (ii), we get u = 1 m/s.

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If body having initial velocity zero is moving with uniform acceleration 8 m/sec² , then the distance travelled by it in fifth second will be

36 metres

40 metres

100 metres

Zero

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Explanation

Distance travelled in $n^{t h}$ second $= u + \frac{a}{2} (2 n - 1)$

Distance travelled in $5^{t h}$ second $= 0 + \frac{8}{2} (2 \times 5 - 1)$ = 36m

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An alpha particle enters a hollow tube of 4 m length with an initial speed of 1 km/s. It is accelerated in the tube and comes out of it with a speed of 9 km/s. The time for which it remains inside the tube is

$8 \times 10^{- 3} s$

$80 \times 10^{- 3} s$

$800 \times 10^{- 3} s$

$8 \times 10^{- 4} s$

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Explanation

$v^{2} = u^{2} + 2 a s \Rightarrow {(9000)}^{2} - {(1000)}^{2} = 2 \times a \times 4$

$\Rightarrow a = 10^{7} m / s^{2}$

Now $t = \frac{v - u}{a}$

$\Rightarrow t = \frac{9000 - 1000}{10^{7}} = 8 \times 10^{- 4} s e c$

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Two cars A and B are travelling in the same direction with velocities v₁ and v₂ $(v_{1} > v_{2})$ . When the car A is at a distance d behind car B, the driver of the car A applied the brake producing a uniform retardation a. There will be no collision when

$d < \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

$d < \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$

$d > \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

$d > \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$

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Explanation

Initial relative velocity $= v_{1} - v_{2}$ , Final relative velocity = 0

From $v^{2} = u^{2} - 2 a s$ ⇒ $0 = {(v_{1} - v_{2})}^{2} - 2 \times a \times s$

⇒ $s = \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

If the distance between two cars is 's' then collision will take place. To avoid collision d > s ∴ $d > \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

where d = actual initial distance between two cars.

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The displacement of a particle is given by $y = a + b t + c t^{2} - d t^{4}$ . The initial velocity and acceleration are respectively

$b, - 4 d$

$- b, 2 c$

$b, 2 c$

$2 c, - 4 d$

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Explanation

$y = a + b t + c t^{2} - d t^{4}$

$∴ v = \frac{d y}{d t} = b + 2 c t - 4 d t^{3}$ and $a = \frac{d v}{d t} = 2 c - 12 d t^{2}$

Hence, at t = 0, v_initial = b and a_initial= 2c.

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Two trains travelling on the same track are approaching each other with equal speeds of 40 m/s. The drivers of the trains begin to decelerate simultaneously when they are just 2.0 km apart. Assuming the decelerations to be uniform and equal, the value of the deceleration to barely avoid collision should be

11.8 m/s²

11.0 m/s²

1.6 m/s²

0.8 m/s²

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Explanation

Both trains will travel a distance of 1 km before to come in rest. In this case by using $v^{2} = u^{2} + 2 a s$

$\Rightarrow 0 = {(40)}^{2} + 2 a \times 1000 \Rightarrow a = - 0 .8 m / s^{2}$

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A body moves from rest with a constant acceleration of 5 m/s². Its instantaneous speed (in m/s) at the end of 10 sec is

0.5

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Explanation

$v = u + a t \Rightarrow v = 0 + 5 \times 10 = 50 m / s$

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A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

Both will be equal

First will be half of second

First will be 1/4 of second

No definite ratio

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Explanation

Let 'a' be the retardation of boggy then distance covered by it be S. If u is the initial velocity of boggy after detaching from train (i.e. uniform speed of train)

$v^{2} = u^{2} + 2 a s \Rightarrow 0 = u^{2} - 2 a s \Rightarrow s_{b} = \frac{u^{2}}{2 a}$

Time taken by boggy to stop

$v = u + a t \Rightarrow 0 = u - a t \Rightarrow t = \frac{u}{a}$

In this time t distance travelled by train $= s_{t} = u t = \frac{u^{2}}{a}$

Hence ratio $\frac{s_{b}}{s_{t}} = \frac{1}{2}$

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A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second

$\frac{7}{5}$

$\frac{5}{7}$

$\frac{7}{3}$

$\frac{3}{7}$

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Explanation

$S_{n} = u + \frac{a}{2} (2 n - 1) = \frac{a}{2} (2 n - 1)$ because $u = 0$

Hence $\frac{S_{4}}{S_{3}} = \frac{7}{5}$

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