Question

A point particle moves along a straight line such that $x=\sqrt{t}$, where $t$ is time. Then, ratio of acceleration to cube of the velocity is

• A. $-1$
• B. $-0.5$
• C. $-3$
• D. $-2$

Answer 1

Answer: (D)

$-2$

Given, $x=\sqrt{t}$

On differentiating, both sides with respect to $t$, we get

$\frac{dx}{dt}=\frac{1}{2\sqrt{t}}=v$ (say)

Again, differentiating both sides with respect to $t$, we get

$\frac{d^{2}x}{d{t}^2}=\frac{1}{2}(-\frac{1}{2}{t}^{-3/2})$

$=-\frac{1}{4t^{3/2}}=a$ (say)

Now, $\frac{a}{v^3}=\frac{-\frac{1}{4t^{3/2}}}{\frac{1}{8t^{3/2}}}$

$=-\frac{1}{4}\times 8=-2$.