JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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Analyzing +4 | -1

The velocity of a particle depends upon the time

\mathrm{t}

according to the equation:

\mathrm{v}=\sqrt{\mathrm{a b}}+\mathrm{b t}+\frac{\mathrm{c}}{\mathrm{d+t}}

. The physical quantities which are represented by

\mathrm{a, b, c}

and

\mathrm{d}

, are in the following order:

View answer and explanation

Correct option: B

Distance, acceleration, distance, time

Option B (Distance, acceleration, distance, time) is correct because dimensional analysis of the given equation reveals the physical quantities represented by a, b, c, and d.

More Questions from MOTION IN A STRAIGHT LINE

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Assertion: In a free fall, the initial velocity of a body may not be zero.
Reason: Free fall means the vertical acceleration of the body is equal to

\mathrm{g}

. A ball is thrown vertically upward with speed 15 m/s and it returns to the ground with speed 10 m/s as a constant air resistance acts. The maximum height attained by the ball is Which of the following is not a derived quantites? Assertion: If a body is dropped from the top of a tower of height

\mathrm{h}

and another body is thrown up simultaneously with velocity

\mathrm{u}

from the foot of the tower, then both of them would meet after a time

\frac{\mathrm{h}}{\mathrm{u}}

.
Reason: For a body projected upwards, the distance covered by the body in the last second of its upward journey is always

4.9 \mathrm{~m}

irrespective of velocity of projection.

(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2)

A particle is having uniform acceleration. If its displacement from

t

(t+1)

second is

120 \mathrm{~m}

and change in velocity is

50 \mathrm{~m} / \mathrm{s}

. Find its displacement (in

\mathrm{m}

) in

(t+2)

second. An aeroplane flies 400m north and 300m south and then flies 1200m upwards, then net displacement is