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

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A particle start moving from rest state along a straight line under the action of a constant force and travel distance x in first 5 seconds. The distance travelled by it in next five seconds will be

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Explanation

Body starts from rest and moves with a constant acceleration, then the distance travelled in equal time intervals
will be in the ratio of odd number. (Galileo's law of odd number)
x1 : x2 ⇒ 1 : 3
x : x2 ⇒ 1 : 3

x = 3x

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The displacement x of a particle along a straight line at time t is given by $x = a_{0} + a_{1} t + a_{2} t^{2}$ . The acceleration of the particle is

$a_{0}$

$a_{1}$

$2 a_{2}$

$a_{2}$

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Explanation

(C)

$v = \frac{d x}{d t} = = a_{1} t^{1 - 1} + 2 a_{2} t^{2 - 1} = a_{1} + 2 a_{2} t a = \frac{d v}{d t} = 2 a_{2} t^{1 - 1} = 2 a_{2}$

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The displacement of a particle is given by $y = a + bt + {ct}^{2} - {dt}^{4}$ . The initial velocity and acceleration are respectively

b, -4d

-d, 2c

b, 2c

2c, -4d

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Explanation

$G i v e n, y = a + b t + c t^{2} - d t^{4} \Rightarrow v = \frac{d v}{d t} = b + 2 c t - 4 d t^{3} \Rightarrow a = \frac{d v}{d t} = 2 c - 12 d t^{2} a t t = 0, v = b a n d a = 2 c$

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A particle moves along a straight line such that its displacement at any time t is given by $s = t^{3} - 3 t^{2} + 6 t - 12$ metres. The velocity when the acceleration is zero is:

$3 {ms}^{- 1}$

$- 12 {ms}^{- 1}$

$42 {ms}^{- 1}$

$- 9 {ms}^{- 1}$

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Explanation

Given,

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

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The position x of particle varies with time t as $x = {at}^{2} - {bt}^{3}$ . The acceleration of the particle will be zero at time equal to

$\frac{a}{b}$

$\frac{2 a}{b}$

$\frac{a}{3 b}$

Zero

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Explanation

Diffrentiating x wrt t ,

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

$v = 2 a t - 3 b t^{2}$

$a = 2 a - 6 b t$

At $t = \frac{a}{3 b} a = 0$

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A particle moves along X-axis in such a way that its X coordinate varies with time t according to the equation $x = (2 - 5 t + 6 t^{2}) m$ . The initial velocity of the particle is

-5 m/s

6 m/s

3 m/s

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Explanation

$v = \frac{d x}{d t} = - 5 + 12 t A t t = 0 v = - 5 m / s$

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A particle is moving along x-axis. The velocity v of particle varies with its position x as $v = \frac{1}{x}$ . Find velocity of particle as a function of time t given that at t=0, x=1 .

$v = \sqrt{2 t + 1}$

$v = \frac{1}{\sqrt{2 t + 1}}$

$v = \frac{1}{t}$

None of these

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Explanation

$v = \frac{1}{x} \frac{d x}{d t} = \frac{1}{x}$

$x d x = d t \int_{1}^{x} x d x = \int_{0}^{t} d t \frac{x^{2} - 1}{2} = t x = \sqrt{2 t + 1} v = \frac{d x}{d t} = \frac{1}{\sqrt{2 t + 1}}$

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A particle moving along a straight line according to the law $x = At + {Bt}^{2} + {Ct}^{3}$ , where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.

Its velocity at t=0 is A

Its acceleration at t=0 is B

Its velocity at t=0 is B

Its acceleration at t=0 is C

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Explanation

$(A) v = \frac{d x}{d t} = A + 2 B t + 3 C t^{2} a = \frac{d v}{d t} = 2 B + 6 C t A t t = 0 v = A a = 2 B$

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If the velocity of a particle moving on x-axis is given by $v = 3 t^{2} - 12 t + 6$ . At which time is the acceleration of particle zero?

$2$ sec

$2 + \sqrt{2}$ sec

$2 - \sqrt{2}$ sec

zero

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Explanation

$(A) a = \frac{d v}{d t} = 6 t - 12 F o r a = 0, 6 t - 12 = 0 t = 2 s e c$

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A particle moves along straight line such that at time t its position from a fixed point O on the line is $x = 3 t^{2} - 2$ . The velocity of the particle when t=2 is:

$8 {ms}^{- 1}$

$4 {ms}^{- 1}$

$12 {ms}^{- 1}$

$0$

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Explanation

(C)

$v = \frac{d x}{d t} = 6 t A t t = 2 s e c v = 12 m / s$

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