I am trying to do the following question for practice (it is an interview question), and I don't know if my methods and answer are correct. Given a velocity $v=2+3t^2$, find acceleration after 3 seconds and distance after 4 seconds. Initial velocity is 0 when $t=0$.
This is how I solved it:
$dv/dt= d(2)/dt + 3 d(t^2)/dt = 0+6t =$ acceleration
acceleration after 3 secs: $18 m/s^2$
acceleration after 4 secs: $24 m/s^2$
velocity after 4 secs: $v= 2+3t^2= 2+48=50$
using equation: $v^2 = u^2 + 2as$
$(50)^2 = 0 + 24*2(s)$