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Motion in a Straight Line NEET Questions — Free MCQ Practice

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A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

A.

$(\frac{α^{2} + β^{2}}{α β}) t$

B.

$(\frac{α^{2} - β^{2}}{α β}) t$

C.

$\frac{(α + β) t}{α β}$

D.

$\frac{α β t}{α + β}$

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Explanation

Let the car accelerate at rate $α$ for time $t_{1}$ then maximum velocity attained, $v = 0 + α t_{1} = α t_{1}$

Now, the car decelerates at a rate $β$ for time $(t - t_{1})$ and finally comes to rest. Then,

$0 = v - β (t - t_{1})$ ⇒ $0 = α t_{1} - β t + β t_{1}$

⇒ $t_{1} = \frac{β}{α + β} t$

∴ $v = \frac{α β}{α + β} t$

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A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t₁ and t₂ seconds respectively, then

A.

$t = t_{1} - t_{2}$

B.

$t = \frac{t_{1} + t_{2}}{2}$

C.

$t = \sqrt{t_{1} t_{2}}$

D.

$t = t_{1}^{2} t_{2}^{2}$

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Explanation

If a stone is dropped from height h

then $h = \frac{1}{2} g t^{2}$ …(i)

If a stone is thrown upward with velocity u then

$h = - u t_{1} + \frac{1}{2} g t_{1}^{2}$ …(ii)

If a stone is thrown downward with velocity u then

$h = u t_{2} + \frac{1}{2} g t_{2}^{2}$ …(iii)

From (i), (ii) and (iii) we get

$- u t_{1} + \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t^{2}$ …(iv)

$u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t^{2}$ …(v)

Dividing (iv) and (v) we get

∴ $\frac{- u t_{1}}{u t_{2}} = \frac{\frac{1}{2} g (t^{2} - t_{1}^{2})}{\frac{1}{2} g (t^{2} - t_{2}^{2})}$

or $- \frac{t_{1}}{t_{2}} = \frac{t^{2} - t_{1}^{2}}{t^{2} - t_{2}^{2}}$

By solving $t = \sqrt{t_{1} t_{2}}$

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A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

A.

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

B.

$\frac{v}{g} [1 - \sqrt{1 + \frac{2 h}{g}}]$

C.

$\frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

D.

$\frac{v}{g} [1 + \sqrt{v^{2} + \frac{2 g}{h}}]$

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Explanation

Since direction of v is opposite to the direction of g and h so from equation of motion

$h = - v t + \frac{1}{2} g t^{2}$

$\Rightarrow g t^{2} - 2 v t - 2 h = 0$

$\Rightarrow t = \frac{2 v \pm \sqrt{4 v^{2} + 8 g h}}{2 g}$

$\Rightarrow t = \frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

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A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be

A.

All equal, being equal to $\sqrt{2 / g}$ second

B.

In the ratio of the square roots of the integers 1, 2, 3.....

C.

In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), (\sqrt{4} - \sqrt{3})$ ....

D.

In the ratio of the reciprocal of the square roots of the integers i.e.,. $\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{4}}$

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Explanation

$h = u t + \frac{1}{2} g t^{2} \Rightarrow 1 = 0 \times t_{1} + \frac{1}{2} g t_{1}^{2} \Rightarrow t_{1} = \sqrt{2 / g}$

Velocity after travelling 1m distance

$v^{2} = u^{2} + 2 g h \Rightarrow v^{2} = {(0)}^{2} + 2 g \times 1 \Rightarrow v = \sqrt{2 g}$

For second 1 meter distance

$1 = \sqrt{2 g} \times t_{2} + \frac{1}{2} g t_{2}^{2} \Rightarrow g t_{2}^{2} + 2 \sqrt{2 g} t_{2} - 2 = 0$

$t_{2} = \frac{- 2 \sqrt{2 g} \pm \sqrt{8 g + 8 g}}{2 g} = \frac{- \sqrt{2} \pm 2}{\sqrt{g}}$

Taking +ve sign $t_{2} = (2 - \sqrt{2}) / \sqrt{g}$

∴ $\frac{t_{1}}{t_{2}} = \frac{\sqrt{2 / g}}{(2 - \sqrt{2}) / \sqrt{g}} = \frac{1}{\sqrt{2} - 1}$ and so on.

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A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g = 9 .8 m / s^{2})$

A.

At least 0.8 m/s

B.

Any speed less than 19.6 m/s

C.

Only with speed 19.6 m/s

D.

More than 19.6 m/s

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Explanation

Interval of ball throw = 2 sec.

If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 sec.

$T > 4$ sec

$\frac{2 u}{g} > 4 s e c \Rightarrow u > 19 .6 m / s$

for u =19.6. First ball will just strike the ground(in sky)

Second ball will be at highest point (in sky)

Third ball will be at point of projection or at ground (not in sky)

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If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

A.

$\frac{1}{2} g t^{2}$

B.

$u t - \frac{1}{2} g t^{2}$

C.

$(u - g t) t$

D.

$u t$

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Explanation

The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t seconds of its downward motion

From $h = u t + \frac{1}{2} g t^{2}$

$h = \frac{1}{2} g t^{2}$ [As u = 0 for it downward motion]

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A car travels a distance S on a straight road in two hours and then returns to the starting point in the next three hours. Its average velocity is

A.

S/5

B.

2S/5

C.

S/2 + S/3

D.

None of the above

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Explanation

Average velocity $= \frac{Total displacement}{Time} = \frac{0}{2 + 3} = 0$

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A body has speed V, 2V and 3V in first 1/3 of distance S, second 1/3 of S and third 1/3 of S respectively. Its average speed will be

A.

V

B.

2V

C.

$\frac{18}{11} V$

D.

$\frac{11}{18} V$

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Explanation

$v_{a v} = \frac{Total distance}{Time taken} = \frac{x}{\frac{x / 3}{v} + \frac{x / 3}{2 v} + \frac{x / 3}{3 v}} = \frac{18}{11} v$

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If the body covers one-third distance at speed v₁, next one third at speed v₂ and last one third at speed v₃, then average speed will be

A.

$\frac{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}{v_{1} + v_{2} + v_{3}}$

B.

$\frac{v_{1} + v_{2} + v_{3}}{3}$

C.

$\frac{v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

D.

$\frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

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Explanation

$v_{a v} = \frac{x}{\frac{x / 3}{v_{1}} + \frac{x / 3}{v_{2}} + \frac{x / 3}{v_{3}}} = \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{1} v_{3}}$

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The displacement of the particle varies with time according to the relation $x = \frac{k}{b} [1 - e^{- b t}]$ . Then the velocity of the particle is

A.

$k (e^{- b t})$

B.

$\frac{k}{b^{2} e^{- b t}}$

C.

$k b e^{- b t}$

D.

None of these

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Explanation

$v = \frac{d x}{d t} = \frac{d}{d t} [\frac{k}{b} (1 - e^{- b t})] = \frac{k}{b} [0 - (- b) e^{- b t}] = k e^{- b t}$ .

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

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