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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: $v^2 = u^2 + 2as$

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

$s= 2500/24*2$

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  • $\begingroup$ what is this equation: $v^2 = u^2 + 2as$? $\endgroup$ – Nikos M. Nov 15 '14 at 22:12
  • $\begingroup$ Can't it be that you made a mistake? Given the velocity formula, at t=0 you should have v=2, not v=0. Probably the exercise means that at t=0 the position is 0. The acceleration you obtain indeed by taking the derivative of the velocity, and the distance is the integral of the velocity over time. HERE you need the position at t=0 to fix the constant of integration. So, the integral over velocity is 2t + t^3 + Const, and if at t=0 the position was 0 (so I guess) then Const = 0 . $\endgroup$ – Sofia Nov 15 '14 at 22:23
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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.

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  • $\begingroup$ Hi mlainz, please note that, on this site, we usually try not to give fully worked solutions to exercises and/or homework-like questions. Don't feel too worried for now, but always keep in mind that one should try not to encourage low-effort homework questions $\endgroup$ – Danu Nov 15 '14 at 23:49
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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).

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