Question

The position of an object varies with the time t as x = at² - bt³. At what time will the acceleration of the object will become zero? (Where a and b are constant)
(A) 2a/3b
(B) a/b
(C) a/3b
(D) zero

Answer 1

Answer:

(C) $\frac{a}{3b}$

Explanation:

as given the position of an object with time =$x=at^{2}-bt^{3}$

velocity ,v is given by rate of change of position ,that is

$v=\frac{\mathrm{d} x}{\mathrm{d} t}$

$v=\frac{\mathrm{d} \left ( at^{2}-bt^{3} \right )}{\mathrm{d} t}$

$v=2at-3bt^{2}$

now acceleration (a) is given bt rate of change of veocity,that is

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=\frac{\mathrm{d} \left ( 2at-3bt^{2} \right )}{\mathrm{d} t}$

$a=2a-6bt$

now put acceleration =0

therefore ,$2a-6bt=0$

                    $t=\frac{a}{3b}$