Blog Log in Get Started Free

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

Practice free Motion in a Straight Line (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

The acceleration of a particle starting from rest, varies with time according to the relation A = – aω² sinω t. The displacement of this particle at a time t will be

$- \frac{1}{2} (a ω^{2} \sin ω t) t^{2}$

$a ω \sin ω t$

$a ω \cos ω t$

$a \sin ω t$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Velocity $v = \int A d t = \int (- a ω^{2} \sin ω t) d t = a ω \cos ω t$

Displacement x $= \int v d t = \int a ω \cos ω t d t = a \sin ω t$

● ● ● ● ●

If the velocity of a particle is (10 + 2t²) m/s, then the average acceleration of the particle between 2s and 5s is

2 m/s²

4 m/s²

12 m/s²

14 m/s²

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Average acceleration $= \frac{Change in velocity}{Time taken} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

$= \frac{[10 + 2 {(5)}^{2}] - [10 + 2 {(2)}^{2}]}{3} = \frac{60 - 18}{3} = 14 m / s^{2}$

● ● ● ● ●

A bullet moving with a velocity of 200 cm/s penetrates a wooden block and comes to rest after traversing 4 cm inside it. What velocity is needed for travelling distance of 9 cm in same block?

100 cm/s

136.2 cm/s

300 cm/s

250 cm/s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

As $v^{2} = u^{2} - 2 a s \Rightarrow u^{2} = 2 a s$ $(∵ v = 0)$

$\Rightarrow u^{2} \propto s$ ⇒ $\frac{u_{2}}{u_{1}} = {(\frac{s_{2}}{s_{1}})}^{1 / 2}$

⇒ $u_{2} = {(\frac{9}{4})}^{1 / 2} u_{1} = \frac{3}{2} u_{1} = 300 c m / s$

● ● ● ● ●

A thief is running away on a straight road in jeep moving with a speed of 9 ms^–1. A police man chases him on a motor cycle moving at a speed of 10 ms^–1. If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take for the police to catch the thief

1 s

19 s

90 s

100 s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The relative velocity of policeman w.r.t. thief = 10 – 9 = 1 m/s.

∴ Time taken by police to catch the thief $= \frac{100}{1}$ =100 sec

● ● ● ● ●

A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h². After how much time will B catch up with A

1 hr

1/2 hr

1/4 hr

1/8 hr

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Let car B catches, car A after ‘t’ sec, then

$60 t + 2 .5 = 70 t - \frac{1}{2} \times 20 \times t^{2}$

⇒ $10 t^{2} - 10 t + 2 .5 = 0$

⇒ $t^{2} - t + 0 .25 = 0$

∴ $t = \frac{1 \pm \sqrt{1 - 4 \times (0 .25)}}{2} = \frac{1}{2} h r$

● ● ● ● ●

The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

$\sqrt{3} u$

$\sqrt{5} u$

$\sqrt{11} u$

$\sqrt{7} u$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

As $v^{2} = u^{2} + 2 a s$ ⇒ ${(2 u)}^{2} = u^{2} + 2 a s \Rightarrow 2 a s = 3 u^{2}$

Now, after covering an additional distance s, if velocity becomes v, then,

$v^{2} = u^{2} + 2 a (2 s) = u^{2} + 4 a s = u^{2} + 6 u^{2} = 7 u^{2}$

∴ $v = \sqrt{7} u$ .

● ● ● ● ●

Two trains one of length 100 m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10 m/s. If their acceleration are 0.3 m/s² and 0.2 m/s² respectively, then the time they take to pass each other will be

5 s

10 s

15 s

20 s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Relative velocity of one train w.r.t. other

= 10 + 10 = 20 m/s.

Relative acceleration =0.3+0.2=0.5 m/s²

If trains cross each other then from $s = u t + \frac{1}{2} a t^{2}$

$A s, s = s_{1} + s_{2} = 100 + 125 = 225$

⇒ $225 = 20 t + \frac{1}{2} \times 0 .5 \times t^{2}$ ⇒ $0 .5 t^{2} + 40 t - 450 = 0$

⇒ $t = - \frac{40 \pm \sqrt{1600 + 4 . (005) \times 450}}{1} = - 40 \pm 50$

∴ t = 10 sec (Taking +ve value).

● ● ● ● ●

A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in the last two seconds is

$\frac{2 v (n + 1)}{n}$

$\frac{v (n + 1)}{n}$

$\frac{v (n - 1)}{n}$

$\frac{2 v (n - 1)}{n}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$∵ v = 0 + n a \Rightarrow a = v / n$

Now, distance travelled in n sec. ⇒ $S_{n} = \frac{1}{2} a n^{2}$ and distance travelled in (n – 2) sec ⇒ $S_{n - 2} = \frac{1}{2} a {(n - 2)}^{2}$

∴ Distance travelled in last two seconds,

$= S_{n} - S_{n - 2} = \frac{1}{2} a n^{2} - \frac{1}{2} a {(n - 2)}^{2}$

$= \frac{a}{2} [n^{2} - {(n - 2)}^{2}] = \frac{a}{2} [n + (n - 2)] [n - (n - 2)]$

= $a (2 n - 2) = \frac{v}{n} (2 n - 2) = \frac{2 v (n - 1)}{n}$

● ● ● ● ●

A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is

$\sqrt{2 t}$

$(2 + \sqrt{2}) t$

$\frac{t}{\sqrt{2}}$

Cannot be predicted unless acceleration is given

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

If a particle starts moving with an acceleration and then experiences an equal retardation, the time taken to return to the initial point is (2 + âˆš2) times the time taken to reach the highest point. This can be derived using kinematic equations and the symmetry of the motion.

● ● ● ● ●

A bird flies for 4 s with a velocity of $| t - 2 | m / s$ in a straight line, where t is time in seconds. It covers a distance of

2 m

4 m

6 m

8 m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Motion in a Straight Line question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.