Question

The position x of a particle varies with time, (t) as $x = at^2 - bt^3.$ The acceleration will be zero at time t is equal to

• A. $\frac{a}{3b}$
• B. zero
• C. $\frac{2a}{3b}$
• D. $\frac{a}{b}$

Answer 1

Answer: (A)

$\frac{a}{3b}$

Answer 2

Distance $(x) = at^2- bt^3.$
Therefore velocity $(v) = \frac{dx}{dt} = \frac{d}{dt} (at^2- bt^3)$
$=2at -3bt^2$ and
acceleration $= \frac{dv}{dt} = \frac{d}{dt} (2at- 3bt^2) = 2a - 6bt = 0$
or $\, \, t = \frac{2a}{6b} = \frac{a}{3b}.$