Question

If a particle starting from rest has an acceleration that increases linearly with time as a=2t, then the distance travelled in third second will be

Answer 1

given that
a=2t
now we know that 
$\frac{dv}{dt} =a$
$dv=2t dt$
$\int\limits^v_0 {} \, dv= \int\limits^t_0 {2t} \, dt$
$v= t^{2}$
now we know that 
$\frac{ds}{dt} =v$$ds= t^{2} dt$
$\int\limits^s_0 {} \, ds= \int\limits^t_0 { t^{2}}\, dt$
$s= \frac{ t^{3} }{3}$
now in three seconds it will travel
$S_{3} = \frac{ 3^{3} }{3}$
$S_{3} = \frac{27}{3}$
in 2 seconds it will travel
$S_{2}= \frac{8}{3}$
therefore in third second it will travel
$S_{3rd}= S_{3} -S_{12}$
$S_{3rd} = \frac{19}{3}$ answer.............

Answer 2

$\frac{dv}{dt} =a=2t$
integrating on both sides
$v=t^2+c$
c is the integration constant
The particle has started from rest 
so at velocity at t=0 is zero
$0=v(0)=c=0$
$v=t^2$
integrating again omn both sides we displacement
$x= \frac{t^3}{2}+C$
where C is an integration constant
the distance travelled in 3rd second
=$=x(3)-x(2)= \frac{27}{3}- \frac{8}{3}= \frac{19}{3}$