Question

The velocity of a particle moving in the positive direction of the x axis varies as $V=\alpha \sqrt{x}$ where $\alpha$ is a positive constant. Assuming that at the moment $t=0$ the particle was located at the point $x=0$, find acceleration at $t=51s$.

• A. ${\alpha }^2$
• B. ${\alpha }^2/2$
• C. $\alpha$
• D. ${\alpha }^3$

Answer 1

The correct option is B ${\alpha }^2/2$

$V=\alpha \sqrt{x}$

We know that $\frac{dV}{dt}=a$

So differentiating both sides wrt $t$ we get

$a=\alpha \frac{1}{2}{x}^{-\frac{1}{2}}$$.\frac{dx}{dt}$, but $\frac{dx}{dt}=v$ so replacing this value in initial equation we get $a=\alpha \frac{1}{2}{x}^{-\frac{1}{2}}$.$\alpha \sqrt{x}$

$=>a=\frac{\alpha ²}{2}$

So best possible option is optionB.