Question

In a straight line motion the distance travelled is proportional to the square root of of the distance, then the acceleration of the particle will be proportional to:

• A. $v^2$
• B. $v^3$
• C. $v^{-1}$
• D. $v$

Answer 1

The correct option is B $v^3$

Given that,

$x\propto t^{\frac{1}{2}}$

Now, by differentiation

$\frac{dx}{dt}\propto \frac{1}{\sqrt{t}}$

$v\propto \frac{1}{\sqrt{t}}$

Now, again differentiation

$\frac{dv}{dt}\propto \frac{1}{t^{\frac{3}{2}}}$

$a\propto \frac{1}{t^{\frac{3}{2}}}$

Now, the acceleration is

$a\propto v^3$

Hence, the acceleration is proportional to $v^3$