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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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Application +4 | -1

A particle moves according to the equation

\frac{d v}{d t}=\alpha-\beta v

, where

\alpha

and

\beta

are constants. Find velocity as a function of time. Assume, body starts from rest

View answer and explanation

Correct option: B

v=(\alpha / \beta)\left(1-e^{-\beta t}\right)

Option B (

v=(\alpha / \beta)(1-e^{-\beta t})

) is correct because it is the solution to the given differential equation with the initial condition that the particle starts from rest.

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)

120 \mathrm{~m}

and change in velocity is

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

. Find its displacement (in

\mathrm{m}

(t+2)

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