Question

A particle is moving on the $x$-axis. The velocity of the particle increases with $x$ as $v=v_0+kx$. Which of the following is true about the acceleration of the particle?

• A. Acceleration is zero
• B. Acceleration is constant but non-zero
• C. Acceleration increases with time
• D. Acceleration decreases with time.

Answer 1

The correct option is C Acceleration increases with time

Differentiate $v=v_o+kx$ with time to get the acceleration
$a=\frac{dv}{dt}=k\frac{dx}{dt}=k(v_o+kx)$
We need to find $x\left(t\right)$ to get time dependence of acceleration $a$.
Re-write given expression of v as $dx=(\frac{v_o}{k}+x)kdt$ and integrate (let the particle start from the origin)
${\int }_o^x\frac{dx}{\frac{v_o}{k}+x}=k{\int }_o^{t}dt$
We get $x=\frac{v_o}{k}(e^{kt}-1)$
Substitute $x$ in the expression of $a$ to get $a=v_{o}k{e}^{kt}$
Hence, we can say that acceleration of the particle increasing with time.