Finding acceleration and distance when velocity given

Question

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:

$v=2+3t^2$

$dv/dt= d(2)/dt + 3 d(t^2)/dt = 0+6t =$ 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: $v^2 = u^2 + 2as$

$(50)^2 = 0 + 24*2(s)$

$s= 2500/24*2$

Answer 1

You should integrate the velocity to get the distance

$$s(4)=x_0-x(4)=\int_{x(0)}^{x(4)}dx=\int_0 ^4vdt=\int_0 ^4( 2 + 3t^2 )dt=2t+t^3|_0^4=2*4+4^3 = 72 \,\text{m}$$

The initial position doesn't matter since de distance $s$ is the difference between the initial position and the position at a given time (in your case, $t=4$).

I don't know what is that formula for. Maybe it only holds for uniform acceleration, (which is not the case).

You got right the rest of the problem.

Answer 2

You made an error. The $v^2=u^2+2aS$ formula is valid only for constant acceleration. In your case, you should integrate velocity from $t=0$ to $t=4$ to get the displacement, which in your case is also the distance(velocity direction does not change).