Question

A particle moves along a straight line in such a way that its acceleration is increasing at the rate of 2 m/s^3. its initial acceleration and velocity is zero the distance covered by it in t=3sec is

Answer 1

da/dt=2m/s^3
da=2dt
by integrating for t=0 to t and a=0 to a.
a=2t
dv/dt=2t
dv=2tdt
integrating for t=0 to t=t and v=0 to v=v.
v=t^2
dx/dt=t^2
integrating for t=0 to t=3sec
x=t^3/3
x=3*3*3/3=9m

Answer 2

Given:

The increase in acceleration is $2m/s^3$

Time t = 3s

Initial velocity =0, initial acceleration = 0

In general, acceleration can be represented as, $\frac { da }{ dt }$ = 2

On integration with respect to t and a, for lower limit t = 0 to t = t and for a = 0 to  a = a

                                                         a = 2t

Acceleration is rate of change of velocity,

                    $\frac { dv }{ dt }$ = 2t

                                                       dv = 2t dt

double integrating with respect to t and v for limits, t = 0 to t and v=0 to v

                                                  v = $t^2$

                             $\frac { dx }{ dt } = t^2$

Integrating for limits t = 0 to t=3

                                  $x = (3)^2 -(0)^2 = 9 m/s$