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

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A 120 m long train is moving in a direction with speed 20 m/s. A train B moving with 30 m/s in the opposite direction and 130 m long crosses the first train in a time

4 s

36 s

38 s

5 s

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Explanation

Total distance $= 130 + 120 = 250 m$

Relative velocity $= 30 - (- 20) = 50 m / s$

Hence $t = 250 / 50 = 5 s$

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A 210 meter long train is moving due North at a speed of 25m/s. A small bird is flying due South a little above the train with speed 5m/s. The time taken by the bird to cross the train is

6 s

7 s

9 s

10 s

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Explanation

Relative velocity of bird w.r.t train $= 25 + 5 = 30 m / s$

time taken by the bird to cross the train $t = \frac{210}{30} = 7 s e c$

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A police jeep is chasing with, velocity of 45 km/h a thief in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is

150 m/s

27 m/s

450 m/s

250 m/s

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Explanation

Effective speed of the bullet

= speed of bullet + speed of police jeep

$= 180 m / s + 45 k m / h = (180 + 12 .5) m / s = 192 .5 m / s$

Speed of thief ’s jeep $= 153 k m / h = 42 .5 m / s$

Velocity of bullet w.r.t thief ’s car $= 192 .5 - 42 .5$ = 150m/s

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A train of 150 meter length is going towards north direction at a speed of 10 m/sec. A parrot flies at the speed of 5 m/sec towards south direction parallel to the railway track. The time taken by the parrot to cross the train is

12 sec

8 sec

15 sec

10 sec

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Explanation

Relative velocity

$= 10 + 5 = 15 m / s e c$

∴ $t = \frac{150}{15} = 10 s e c$

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A boat is moving with velocity of $3 \hat{i} + 4 \hat{j}$ in river and water is moving with a velocity of $- 3 \hat{i} - 4 \hat{j}$ with respect to ground. Relative velocity of boat with respect to water is :

$- 6 \hat{i} - 8 \hat{j}$

$6 \hat{i} + 8 \hat{j}$

$8 \hat{j}$

$6 \hat{i}$

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Explanation

The relative velocity of boat w.r.t. water

$= v_{b o a t} - v_{w a t e r} = (3 \hat{i} + 4 \hat{j}) - (- 3 \hat{i} - 4 \hat{j}) = 6 \hat{i} + 8 \hat{j}$

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The distance between two particles is decreasing at the rate of 6 m/sec. If these particles travel with same initial speeds and in the same direction, then the separation increases at the rate of 4 m/sec. The particles have speeds as

5 m/sec ; 1 m/sec

4 m/sec ; 1 m/sec

4 m/sec ; 2 m/sec

5 m/sec ; 2 m/sec

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Explanation

When two particles moves towards each other then $v_{1} + v_{2} = 6$ ...(i)

When these particles moves in the same direction then $v_{1} - v_{2} = 4$ ...(ii)

By solving $v_{1} = 5 m / s$ and $v_{2} = 1 m / s$

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An express train is moving with a velocity v₁. Its driver finds another train is moving on the same track in the same direction with velocity v₂. To escape collision, driver applies a retardation a on the train. The minimum time of escaping collision will be

$t = \frac{v_{1} - v_{2}}{a}$

$t_{1} = \frac{v_{1}^{2} - v_{2}^{2}}{2}$

None

Both

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Explanation

As the trains are moving in the same direction. So the initial relative speed $(v_{1} - v_{2})$ and by applying retardation final relative speed becomes zero.

From $v = u - a t$ ⇒ $0 = (v_{1} - v_{2}) - a t$ ⇒ $t = \frac{v_{1} - v_{2}}{a}$

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A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The displacement of the stone from the point of release after 10 sec is

490 m

510 m

610 m

725 m

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Explanation

$u = 12 m / s$ , $g = 9 .8 m / \sec^{2}$ , $t = 10 \sec$

Displacement $= u t + \frac{1}{2} g t^{2}$

$= 12 \times 10 + \frac{1}{2} \times 9 .8 \times 100 = 610 m$

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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

2100 m/sec² downwards

2100 m/sec² upwards

1400 m/sec²

700 m/sec²

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Explanation

Velocity at the time of striking the floor,

$u = \sqrt{2 g h_{1}} = \sqrt{2 \times 9 .8 \times 10} = 14 m / s$

Velocity with which it rebounds.

$v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9 .8 \times 2 .5} = 7 m / s$

∴ Change in velocity $Δ v = 7 - (- 14) = 21 m / s$

∴ Acceleration $= \frac{Δ v}{Δ t} = \frac{21}{0 .01} = 2100 m / s^{2}$ (upwards)

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A body A is projected upwards with a velocity of 98 m/s. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

6 sec

8 sec

10 sec

12 sec

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Explanation

Let t be the time of flight of the first body after meeting, then $(t - 4)$ sec will be the time of flight of the second body. Since $h_{1} = h_{2}$

∴ $98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g {(t - 4)}^{2}$

On solving, we get $t = 12$ seconds

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