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Motion in a Straight Line NEET Questions — Free MCQ Practice

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A body A starts from rest with an acceleration a₁. After 2 seconds, another body B starts from rest with an acceleration a₂. If they travel equal distances in the 5th second, after the start of A, then the ratio a₁ : a₂ is equal to

A.

5 : 9

B.

5 : 7

C.

9 : 5

D.

9 : 7

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Explanation

According to problem

Distance travelled by body A in 5^th sec and distance travelled by body B in 3^rdsec. of its motion are equal.

$0 + \frac{a_{1}}{2} (2 \times 5 - 1) = 0 + \frac{a_{2}}{2} [2 \times 3 - 1]$

$9 a_{1} = 5 a_{2} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{5}{9}$

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The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be

A.

$10 \times 10^{4}$ m/s²

B.

$12 \times 10^{4}$ m/s²

C.

$13 .5 \times 10^{4}$ m/s²

D.

$15 \times 10^{4}$ m/s²

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Explanation

$u = 200 m / s, v = 100 m / s, s = 0 .1 m$

$a = \frac{u^{2} - v^{2}}{2 s} = \frac{{(200)}^{2} - {(100)}^{2}}{2 \times 0 .1} = 15 \times 10^{4} m / s^{2}$

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A particle starts from rest, accelerates at 2 m/s² for 10s and then goes for constant speed for 30s and then decelerates at 4 m/s² till it stops. What is the distance travelled by it ?

A.

750 m

B.

800 m

C.

700 m

D.

850 m

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Explanation

Velocity acquired by body in 10sec

$v = 0 + 2 \times 10 = 20 m / s$

and distance travelled by it in 10 sec

$S_{1} = \frac{1}{2} \times 2 \times {(10)}^{2} = 100 m$

then it moves with constant velocity (20 m/s) for 30 sec

$S_{2} = 20 \times 30 = 600 m$

After that due to retardation $(4 m / s^{2})$ it stops

$S_{3} = \frac{v^{2}}{2 a} = \frac{{(20)}^{2}}{2 \times 4} = 50 m$

Total distance travelled $S_{1} + S_{2} + S_{3} = 750 m$

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The engine of a motorcycle can produce a maximum acceleration 5 m/s². Its brakes can produce a maximum retardation 10 m/s². What is the minimum time in which it can cover a distance of 1.5 km

A.

30 sec

B.

15 sec

C.

10 sec

D.

5 sec

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A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

A.

6m

B.

12m

C.

18m

D.

24m

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Explanation

$S \propto u^{2}$ . Now speed is two times so distance will be four times $S = 4 \times 6 = 24 m$

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A student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of 1ms^–2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is

A.

5 ms^–1

B.

8 ms^–1

C.

10 ms^–1

D.

12 ms^–1

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Explanation

Let student will catch the bus after t sec. So it will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2} a t^{2}$ .

For the given condition;

$u t = 50 + \frac{1}{2} a t^{2} = 50 + \frac{t^{2}}{2}$ [ $a = 1 m / s^{2}$ ]

⇒ u $= \frac{50}{t} + \frac{t}{2}$

To find the minimum value of u

$\frac{d u}{d t} = 0$ , so we get $t = 10 \sec$ , then $u = 10 m / s$

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A body A moves with a uniform acceleration a and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity v. The two bodies meet after a time t. The value of t is

A.

$\frac{2 v}{a}$

B.

$\frac{v}{a}$

C.

$\frac{v}{2 a}$

D.

$\sqrt{\frac{v}{2 a}}$

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Explanation

$\frac{1}{2} a t^{2} = v t \Rightarrow t = \frac{2 v}{a}$

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

A.

–5 m/s

B.

6 m/s

C.

–3 m/s

D.

3 m/s

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Explanation

The velocity of the particle is

$\frac{d x}{d t} = \frac{d}{d t} (2 - 5 t + 6 t^{2}) = (0 - 5 + 12 t)$

For initial velocity t = 0, hence $v = - 5 m / s$ .

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A car starts from rest and moves with uniform acceleration a on a straight road from time t = 0 to t = T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is

A.

$\frac{a T}{4}$

B.

$\frac{3 a T}{2}$

C.

$\frac{a T}{2}$

D.

aT

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Explanation

For First part,

u = 0, t = T and acceleration = a

$∴ v = 0 + a T = a T$ and $S_{1} = 0 + \frac{1}{2} a T^{2} = \frac{1}{2} a T^{2}$

For Second part,

$u = a T,$ retardation=a₁, $v = 0$ and time taken = T₁ (let)

∴ $0 = u - a_{1} T_{1} \Rightarrow a T = a_{1} T_{1}$

and from $v^{2} = u^{2} - 2 a S_{2} \Rightarrow S_{2} = \frac{u^{2}}{2 a_{1}} = \frac{1}{2} \frac{a^{2} T^{2}}{a_{1}}$

$S_{2} = \frac{1}{2} a T \times T_{1}$ $(A s a_{1} = \frac{a T}{T_{1}})$

∴ $v_{a v} = \frac{S_{1} + S_{2}}{T + T_{1}} = \frac{\frac{1}{2} a T^{2} + \frac{1}{2} a T \times T_{1}}{T + T_{1}}$

$= \frac{\frac{1}{2} a T (T + T_{1})}{T + T_{1}} = \frac{1}{2} a T$

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An object accelerates from rest to a velocity 27.5 m/s in 10 sec .Then find distance covered by object in next 10 sec

A.

550 m

B.

137.5 m

C.

412.5 m

D.

275 m

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Explanation

u = 0, $v = 27 .5 m / s$ and t = 10 sec

$∴ a = \frac{27 .5 - 0}{10} = 2 .75 m / s^{2}$

Now, the distance traveled in next 10 sec,

$S = u t + \frac{1}{2} a t^{2} = 27 .5 \times 10 + \frac{1}{2} \times 2 .75 \times 100$

= 275 + 137.5 = 412.5 m

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

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