For motion of an object along $x$-axis, the velocity $v$ depends on the position $x$ as $v=\left(2 x^2-x\right) \mathrm{m} / \mathrm{s}$, where $x$ is in $\mathrm{m}$. The acceleration of the object at $x=1 \mathrm{~m}$ will be

JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

1 Medium

1 of 1

Application +4 | -1

For motion of an object along

x

-axis, the velocity

v

depends on the position

x

v=\left(2 x^2-x\right) \mathrm{m} / \mathrm{s}

, where

x

is in

\mathrm{m}

. The acceleration of the object at

x=1 \mathrm{~m}

will be

View answer and explanation

Correct option: B

3 \mathrm{~m} / \mathrm{s}^2

Option B (

3 \mathrm{~m} / \mathrm{s}^2

) is correct because it is the result of applying the chain rule to find acceleration as

a = v(dv/dx)

and evaluating at

x=1 \mathrm{~m}

Practice All Chapter MCQs →

\mathrm{g}

\mathrm{h}

\mathrm{u}

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

4.9 \mathrm{~m}

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

t

(t+1)

120 \mathrm{~m}

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

\mathrm{m}

(t+2)