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

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The acceleration ‘a’ in m/s² of a particle is given by $a = 3 t^{2} + 2 t + 2$ where t is the time. If the particle starts out with a velocity u = 2 m/s at t = 0, then the velocity at the end of 2 second is

12 m/s

18 m/s

27 m/s

36 m/s

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Explanation

$v = u + \int_{}^{} a d t = u + \int_{}^{} (3 t^{2} + 2 t + 2) d t$

$= u + \frac{3 t^{3}}{3} + \frac{2 t^{2}}{2} + 2 t = u + t^{3} + t^{2} + 2 t$

$= 2 + 8 + 4 + 4 = 18 m / s$ (As t = 2 sec)

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A particle moves along a straight line such that its displacement at any time t is given by $S = t^{3} - 6 t^{2} + 3 t + 4$ metres

The velocity when the acceleration is zero is

4 $m s^{- 1}$

$- 12 m s^{- 1}$

$42 m s^{- 1}$

$- 9 m s^{- 1}$

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Explanation

$v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$ and $a = \frac{d v}{d t} = 6 t - 12$

For $a = 0$ , we have $t = 2$ and at $t = 2, v = - 9 m s^{- 1}$

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If a body starts from rest and travels 120 cm in the 6^th second, then what is the acceleration

0.20 m/s²

0.027 m/s²

0.218 m/s²

0.03 m/s²

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Explanation

$S_{n} = u + \frac{a}{2} (2 n - 1)$ ⇒ $1 .2 = 0 + \frac{a}{2} (2 \times 6 - 1)$

⇒ $a = \frac{1 .2 \times 2}{11} = 0 .218 m / s^{2}$

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If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of

20 m

400 m

1440 m

2880 m

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Explanation

Here $v = 144 k m / h = 40 m / s$

$v = u + a t$ ⇒ $40 = 0 + 20 \times a \Rightarrow a = 2 m / s^{2}$

∴ $s = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times {(20)}^{2} = 400 m$

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If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

20 ms^–2

10 ms^–2

2 ms^–2

1 ms^–2

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Explanation

$u = 72 k m p h = 20 m / s,$

By using $v^{2} = u^{2} - 2 a s$ ⇒ $a = \frac{u^{2}}{2 s} = \frac{{(20)}^{2}}{2 \times 200} = 1 m / s^{2}$

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The displacement of a particle starting from rest (at t = 0) is given by $s = 6 t^{2} - t^{3}$ . The time in seconds at which the particle will attain zero velocity again, is

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Explanation

$v = \frac{d s}{d t} = 12 t - 3 t^{2}$

Velocity is zero for $t = 0$ and $t = 4 \sec$

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Two cars A and B are at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of 4 m/s², then B will catch A after how much time ?

10 sec

20 sec

30 sec

35 sec

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Explanation

Let A and B will meet after time t sec. it means the distance travelled by both will be equal.

$S_{A} = u t = 40 t$ and $S_{B} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 4 \times t^{2}$

$S_{A} = S_{B} \Rightarrow 40 t = \frac{1}{2} 4 t^{2} \Rightarrow t = 20$ sec

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The motion of a particle is described by the equation $x = a + b t^{2}$ where a = 15 cm and b = 3 cm/s². Its instantaneous velocity at time 3 sec will be

36 cm/sec

18 cm/sec

16 cm/sec

32 cm/sec

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Explanation

$x = a + b t^{2}, v = \frac{d x}{d t} = 2 b t$

Instantaneous velocity $v = 2 \times 3 \times 3 = 18 c m / \sec$

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A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S₁, S₂ and S₃ in the first five seconds, second five seconds and next five seconds respectively the relation between S₁, S₂ and S₃ is

$S_{1} = S_{2} = S_{3}$

$5 S_{1} = 3 S_{2} = S_{3}$

$S_{1} = \frac{1}{3} S_{2} = \frac{1}{5} S_{3}$

$S_{1} = \frac{1}{5} S_{2} = \frac{1}{3} S_{3}$

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Explanation

Distance travelled in first 5 sec $S_{1} = \frac{1}{2} a (5)^{2}$

Distance travelled in next 5 sec $S_{2} = \frac{1}{2} a (10)^{2} - \frac{1}{2} a (5)^{2} = 3 \frac{a (5)^{2}}{2}$

Distance travelled from t =11s to t=15 s,

$S_{3} = \frac{a (15)^{2}}{2} - \frac{a (10)^{2}}{2} = \frac{5 a (5)^{2}}{2}$

If the body starts from rest and moves with constant acceleration then the ratio of distances in consecutive equal time interval $S_{1} : S_{2} : S_{3} = 1 : 3 : 5$

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A body is moving according to the equation $x = a t + b t^{2} - c t^{3}$ where x = displacement and a, b and c are constants. The acceleration of the body is

$a + 2 b t$

$2 b + 6 c t$

$2 b - 6 c t$

$3 b - 6 c t^{2}$

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Explanation

$x = a t + b t^{2} - c t^{3}, a = \frac{d^{2} x}{d t^{2}} = 2 b - 6 c t$

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