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

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If the distance 's' travelled by a body in time 't' is given by $s = \frac{a}{t} + {bt}^{2}$ then the acceleration equals

$\frac{2 a}{t^{3}} + 2 b$

$\frac{2 s}{t^{3}}$

$2 b - \frac{2 a}{t^{3}}$

$\frac{s}{t^{2}}$

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Explanation

(A)

$s = \frac{a}{t} + {bt}^{2} v = \frac{ds}{dt} = \frac{- a}{t^{2}} + 2 bt a = \frac{dv}{dt} = \frac{2 a}{t^{3}} + 2 b$

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The velocity of a particle moving on the x-axis is given by $v = x^{2} + x$ where v is in m/s and x is in m. Find its acceleration in $m / s^{2}$ when passing through the point x=2m.

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Explanation

$v = x^{2} + x Differentiating both sides w . r . t t \frac{dv}{dt} = \frac{d (x^{2})}{dt} + \frac{dx}{dt} a = 2 x (\frac{dx}{dt}) + \frac{dx}{dt} a = (2 x + 1) v [∵ v = \frac{dx}{dt}] = (2 x + 1) (x^{2} + x) At x = 2 m a = [2 (2) + 1] [2^{2} + 2] = 30 {ms}^{- 2}$

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A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial velocity of the particle is

1 m/s

3 m/s

12 m/s

20 m/s

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Explanation

$v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 A t t = 0 v = 12 m / s$

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A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial acceleration of the particle is

Zero

$1 m / s^{2}$

$- 3 m / s^{2}$

$- 6 m / s^{2}$

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Explanation

$x = t^{3} - 3 t^{2} + 12 t + 20 v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 a = \frac{d v}{d t} = 6 t - 6 a t t = 0, a = - 6 m / s^{2}$

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A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Velocity of the particle when its acceleration zero is

1 m/s

3 m/s

6 m/s

9 m/s

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Explanation

$v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 a = \frac{d v}{d t} = 6 t - 6 W h e n a = 0, 6 t - 6 = 0 \Rightarrow t = 1 s e c v a t 1 s e c . = 9 m / s$

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The acceleration of a particle starting from rest varies with time according to relation, $a = α t + β$ . Find the velocity of the particle at time instant t.

$α t^{2} + β t$

$α t^{2} + \frac{β t}{2}$

$\frac{α t^{2}}{2} + β t$

$2 α t^{2} + β t$

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Explanation

Given that, $a = \frac{d v}{d t} = α t + β$

${}_{0}^{v}\int d v = {}_{0}^{t}\int (α t + β) d t = {[\frac{α t^{2}}{2} + β t]}_{0}^{t} \Rightarrow v = \frac{α t^{2}}{2} + β t$

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The velocity of a body moving with a uniform acceleration of 2 m/sec² is 10 m/sec. Its velocity after an interval of 4 sec is

12 m/sec

14 m/sec

16 m/sec

18 m/sec

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Explanation

$v = u + a t = 10 + 2 \times 4 = 18 m / \sec$

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A particle starting from rest moving with constant acceleration travels a distance x in first 2 seconds and a distance y in next two seconds, then

y = x

y = 2x

y = 3x

y = 4x

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Explanation

$Dist . covered in 1^{st} 2 \sec : ‐ x = o + \frac{1}{2} a \times {(2)}^{2} = 2 a Dist . covered in 4 \sec : ‐ d = o + \frac{1}{2} \times a \times {(4)}^{2} = 8 a \Rightarrow y = d - x = 8 a - 2 a = 6 a \Rightarrow \frac{x}{y} = \frac{2 a}{6 a} \Rightarrow y = 3 x$

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The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s². The distance covered by the body in the 5^th second of its motion is

25 m

35 m

50 m

85 m

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Explanation

$S_{n} = u + \frac{a}{2} [2 n - 1]$

$S_{5^{t h}} = 7 + \frac{4}{2} [2 \times 5 - 1] = 7 + 18 = 25 m$ .

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The velocity of a body depends on time according to the equation $v = 20 + 0 .1 t^{2}$ . The body is undergoing

Uniform acceleration

Uniform retardation

Non-uniform acceleration

Zero acceleration

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Explanation

Acceleration $a = \frac{d v}{d t} = 0 .1 \times 2 t = 0 .2 t$

Which is time dependent i.e. non-uniform acceleration.

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