Question

A particle moves along a straight line in such a way that its acceleration is increasing at the rate of $2m/s^3$. Its initial acceleration and velocity were zero. The distance which it covers in the $3^{rd}$ second is $x$ meter. Find $3x$.

Answer 1

Given:
$\frac{da}{dt}=2da=2dt$

Integrating both sides,
${\int }_0^{a}da={\int }_0^{t}dta=2t...\left(i\right)$

As we know,
$a=\frac{dv}{dt}$

Therefore, from $\left(i\right)$m equation we get,
$adt=dv$
Integrating both sides,
${\int }_0^{v}dv={\int }_0^{t}2tdtv=t^2...\left(ii\right)$

As we know,
$v=\frac{dx}{dt}....\left(iii\right)$
Putting the value of $v$ in equation (iii)
we get,
${\int }_0^{x}dx={\int }_2^3{t}^{2}dtx=\frac{t^3}{3}{|}_2^3=\frac{27}{3}-\frac{8}{3}x=\frac{19}{3}m$

Hence,
$3x=3\times \frac{19}{3}=19m$