Question

If the velocity of a particle is given by
$v=(180-16x{)}^{\frac{1}{2}}$ $m$/$s$, then its acceleration will be

• A. $8m/s^2$
• B. Zero
• C. $-8m/s^2$
• D. $4m/s^2$

Answer 1

The correct option is C $-8m/s^2$

Given, speed of the particle,

$v=(180-16x{)}^{\frac{1}{2}}$

So, acceleration of the particle,

$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$

$a=(180-16x{)}^{\frac{1}{2}}$ $\left(\frac{1}{2}\right(180-16x{)}^{\frac{-1}{2}}(-16))$

$a=-8m/s^2$