For a $1-D$ motion, relation between position $x$ and time $t$ is $t=\alpha x^2+\beta x$. Find the relation between velocity $v$ and acceleration $a$.

JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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For a

1-D

motion, relation between position

x

and time

t

t=\alpha x^2+\beta x

. Find the relation between velocity

v

and acceleration

a

View answer and explanation

Correct option: C

a=-2 \alpha v^3

Option C (

a=-2 \alpha v^3

) is correct because it is derived by differentiating the given equation with respect to time and using the chain rule to relate acceleration and velocity.

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

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. Find its displacement (in

\mathrm{m}

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(t+2)

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