Experiencing Jerk

da/dt=2

da=2dt

integrating from 0 to t, we get

a=2t

dv/dt=2t (as a=dv/dt)

dv=2tdt

integrating

v=t*t

dx/dt=t*t (as v=dx/dt)

dx=t*t dt

integrating from o to t

x=(t * t * t)/3

when t=6

thus x= 72

a=2t then v=t^2 then s=t^3/3 then integrate 0 to 6 s=72

$Given$ $that$ $\frac{da}{dt}$ = $2$ or, $da$ = $2dt$ or, $\int\ da$ = $\int\ 2dt$ or, $a$ = $2t$ or, $\frac{dv}{dt}$ = $2t$ or, $dv$ = $2tdt$ or, $\int\ dv$ = $\int\ 2tdt$ or, $v$ = $t^2$ or, $\frac{dx}{dt}$ = $t^2$ or, $dx$ = $t^2$ $dt$ or, $\int\ dx$ = $\int\ t^2dt$ or, $x$ = $\frac{t^3}{3}$ $now$ $plugging$ $the$ $limits$ $0$ $to$ $6$ $we$ $get$ $x$ = $\frac{6^3}{3}$ - $\frac{0^3}{3}$ $x$ = $72$

$so$ $answer$ $is$ $\boxed {72}$

Guys, there is a mistake in this question.its actually unsolvable without an initial condition for acceleration."initially at rest " simply implies v(t=0)=0 not a(t=0)=0.Therefor upon first integration we gets a=2t+C where C is a constant and finally S=(t^3)/3+C(t^2)/2.I solved it by assuming C=0.actually one more initial condition was needed.