Question

If the velocity of a particle is given by $v=(18016x{)}^{\frac{1}{2}}m/s$, then the acceleration is

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

Answer 1

Answer: (C)

$-8m/s^2$

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

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

Diff. both sides wrt. x, we get:
$2v\frac{dv}{dx}=-16v\frac{dv}{dx}=a=-8m/s^2$