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

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A particle is projected with velocity $v_{0}$ along x-axis. The deceleration of the particle is proportional to the square of the distance from the origin i.e., $a = α x^{2} .$ The distance at which the particle stops is :

$\sqrt{\frac{3 v_{0}}{2 α}}$

${(\frac{3 v_{o}}{2 α})}^{\frac{1}{3}}$

$\sqrt{\frac{3 v {}_{0}^{2}}{2 α}}$

${(\frac{3 v {}_{0}^{2}}{2 α})}^{\frac{1}{3}}$

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Explanation

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = - α x^{2}$ (given)

⇒ $\int_{v_{0}}^{0} v d v = - α \int_{0}^{S} x^{2} d x$ ⇒ ${[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{S}$

⇒ $\frac{v_{0}^{2}}{2} = \frac{α S^{3}}{3}$ ⇒ $S = {[\frac{3 v_{0}^{2}}{2 α}]}^{\frac{1}{3}}$

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A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

$\vec{v} / 2$ and v/2

0 and v/2

0 and 0

$\vec{v} / 2$ and 0

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Explanation

Average velocity = 0 because net displacement of the body is zero.

Average speed $= \frac{Total distance covered}{Time of flight} = \frac{2 H_{\max}}{2 u / g}$

⇒ $v_{a v} = \frac{2 u^{2} / 2 g}{2 u / g}$ ⇒ $v_{a v} = u / 2$

Velocity of projection = v (given)

∴ $v_{av} = v / 2$

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A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after time

$\sqrt{\frac{h}{8 g}}$

$\sqrt{8 g h}$

$\sqrt{2 g h}$

$\sqrt{\frac{h}{2 g}}$

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Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third and the third and forth will be respectively

35, 25 and 15 m

30, 20 and 10 m

20, 10 and 5 m

40, 30 and 20 m

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A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

$4 \sqrt{\frac{h}{g}}$

$2 \sqrt{\frac{h}{g}}$

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{g}{h}}$

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Explanation

The velocity of balloon at height h, $v = \sqrt{2 (\frac{g}{8}) h}$

When the stone released from this balloon, it will go upward with velocity v = $\frac{\sqrt{g h}}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground using 2nd equation of motion:

$- h = \frac{\sqrt{g h}}{2} t - \frac{1}{2} g t^{2} t^{2} - \sqrt{\frac{h}{g}} t - \frac{2 h}{g} = 0 t = \frac{\sqrt{\frac{h}{g}} \pm \sqrt{\frac{h}{g} + 4 \times \frac{2 h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} \pm 3 \sqrt{\frac{h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} + 3 \sqrt{\frac{h}{g}}}{2} = 2 \sqrt{\frac{h}{g}}$

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A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level. The height of the tower is

50 m

75 m

100 m

125 m

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Explanation

Let height of tower is h and body takes t time to reach to ground when it fall freely.

∴ $h = \frac{1}{2} g t^{2}$ …(i)

In last second i.e. t^{^th} sec body travels = 0.36 h

It means in rest of the time i.e. in (t – 1) sec it travels

$= h - 0 .36 h = 0 .64 h$

Now applying equation of motion for (t – 1) sec

$0 .64 h = \frac{1}{2} g {(t - 1)}^{2}$ …(ii)

From (i) and (ii) we get, t = 5 sec and h = 125 m

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Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_{p} (t) = a t + b t^{2}$ and $x_{Q} (t) = f t - t^{2}$ . At what time do the cars have the same velocity?

$\frac{a - f}{1 + b}$

$\frac{a + f}{2 (b - 1)}$

$\frac{a + f}{2 (1 + b)}$

$\frac{f - a}{2 (1 + b)}$

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Explanation

(d) Velocity of each car is given by -

$v_{p} = \frac{d x (t)}{d t} = a + 2 b t$

$And, v_{Q} = \frac{{dx}_{Q} (t)}{dt} = f - 2 t$

$It is given that - v_{P} = v_{Q} \Rightarrow a + 2 bt = f - 2 t \Rightarrow t = \frac{f - a}{2 (b + 1)}$

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The motion of a particle along a straight line is described by equation

$x = 8 + 12 t - t^{3}$

where x is in metre and t is in second. The retardation of the particle when its velocity becomes zero is

24 m $s^{- 2}$

zero

6 $m s^{- 2}$

12 $m s^{- 2}$

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Explanation

Given, $x = 8 + 12 t - t^{3}$

We know $v = \frac{d x}{d t}$

So, $v = 12 - 3 t^{2}$

When v=0, t=2 sec

and $a = \frac{d v}{d t}$

and a=-6t

At t=2s,

$a = - 12 m / s^{2}$

So, retardation of the particle = 12m/ $s^{2} .$

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A particle covers half of its total distance with speed $v_{1}$ and the rest half distance with speed $v_{2} .$ Its average speed during the complete journey is

$\frac{v_{1} v_{2}}{v_{1} + v_{2}}$

$\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

$\frac{2 v_{1}^{2} v_{2}^{2}}{v_{1}^{2} + v_{2}^{2}}$

$\frac{v_{1} + v_{2}}{2}$

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Explanation

Velocity $v = \frac{s}{t} \Rightarrow t = \frac{s}{v}$

The average speed of the particle :

$v_{av} = \frac{s + s}{\frac{s}{v_{1}} + \frac{s}{v_{2}}} v_{av} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

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A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is $s_{1}$ and that covered in the first 20 s is $s_{2},$ then

$s_{2} = 2 s_{1}$

$s_{2} = 3 s_{1}$

$s_{2} = 4 s_{1}$

$s_{2} = s_{1}$

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Explanation

(c)

Key Idea If the particle is moving in a straight line under the action of a constant force then

distance covered $s = ut + \frac{1}{2} {at}^{2} .$

Since the body starts from rest u=0

$∴ s = \frac{1}{2} {at}^{2} Now, s_{1} = \frac{1}{2} a {(10)}^{2} . . . . (i) and s_{2} = \frac{1}{2} a (20)^{2} . . . . (ii) Dividing Eq . (i) and Eq . (ii), we get \frac{s_{1}}{s_{2}} = \frac{(10)^{2}}{(20)^{2}} \Rightarrow s_{2} = 4 s_{1}$

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