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Motion in a Lane MCQs for NEET — Physics Questions with Answers

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The x and y coordinates of a particle at any time t are given by $x = 7 t + 4 t^{2}$ and y = 5t, where x and y are in metre and t in seconds. The acceleration of particle at t = 5s is

Zero

8 m/s²

20 m/s²

40 m/s²

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Explanation

$a = \sqrt{a_{x}^{2} + a_{y}^{2}} = {[{(\frac{d^{2} x}{d t^{2}})}^{2} + {(\frac{d^{2} y}{d t^{2}})}^{2}]}^{\frac{1}{2}}$

Here $\frac{d^{2} y}{d t^{2}} = 0$ . Hence $a = \frac{d^{2} x}{d t^{2}} = 8 m / s^{2}$

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A boat is sent across a river in perpendicular direction with a velocity of 8 km/hr. If the resultant velocity of boat is 10 km/hr, then velocity of the river is :

10 km/hr

8 km/hr

6 km/hr

4 km/hr

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The coordinates of a moving particle at any time ‘t’ are given by x = αt³ and y = βt³. The speed of the particle at time ‘t’ is given by

$\sqrt{α^{2} + β^{2}}$

$3 t \sqrt{α^{2} + β^{2}}$

$3 t^{2} \sqrt{α^{2} + β^{2}}$

$t^{2} \sqrt{α^{2} + β^{2}}$

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Explanation

$x = α t^{3}$ and $y = β t^{3}$ (given)

$v_{x} = \frac{d x}{d t} = 3 α t^{2}$ and $v_{y} = \frac{d y}{d t} = 3 β t^{2}$

Resultant velocity $= v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 3 t^{2} \sqrt{α^{2} + β^{2}}$

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A stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following

Straight path

Circular path

Parabolic path

Hyperbolic path

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Explanation

Due to constant velocity along horizontal and vertical downward force of gravity stone will hit the ground following parabolic path.

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An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth's surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling

On a parabolic path as seen by pilot in the plane

Vertically along a straight path as seen by an observer on the ground near the target

On a parabolic path as seen by an observer on the ground near the target

On a zig-zag path as seen by pilot in the plane

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Explanation

The pilot will see the ball falling in straight line because the reference frame is moving with the same horizontal velocity but the observer at rest will see the ball falling in parabolic path.

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A bomb is dropped from an aeroplane moving horizontally at constant speed. When air resistance is taken into consideration, the bomb

Falls to earth exactly below the aeroplane

Fall to earth behind the aeroplane

Falls to earth ahead of the aeroplane

Flies with the aeroplane

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Explanation

Due to air resistance, it’s horizontal velocity will decrease so it will fall behind the aeroplane.

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An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is

1200 m

0.33 km

3.33 km

33 km

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Explanation

Horizontal displacement of the bomb

AB = Horizontal velocity × time available

$A B = u \times \sqrt{\frac{2 h}{g}} = 600 \times \frac{5}{18} \times \sqrt{\frac{2 \times 1960}{9 .8}}$ = 3.33 Km

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An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/sec²)

3 sec and 2000 m

5 sec and 500 m

8 sec and 1500 m

9 sec and 1800 m

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Explanation

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 396 .9}{9 .8}} ≃ 9 \sec$ and $u=720 km/hr=200 m/s$

∴ $R = u \times t = 200 \times 9 = 1800 m$

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A particle (A) is dropped from a height and another particle (B) is thrown in horizontal direction with speed of 5 m/sec from the same height. The correct statement is

Both particles will reach at ground simultaneously

Both particles will reach at ground with same speed

Particle (A) will reach at ground first with respect to particle (B)

Particle (B) will reach at ground first with respect to particle (A)

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Explanation

For both cases $t = \sqrt{\frac{2 h}{g}} =$ constant.

Because vertical downward component of velocity will be zero for both the particles.

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A bomber plane moves horizontally with a speed of 500 m/s and a bomb is released from it. The bomb strikes the ground in 10 sec. Angle at which it strikes the ground will be (g = 10 m/s²)

$\tan^{- 1} (\frac{1}{5})$

$\tan (\frac{1}{5})$

tan^–1(1)

tan^–1(5)

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