Calculate acceleration and time given initial speed, final speed, and travelling distance? [closed]

A motorcycle is known to accelerate from rest to 190km/h in 402m.

Considering the rate of acceleration is constant, how should I go about calculating the acceleration rate and the time it took the bike to complete the distance?

closed as too localized by Qmechanic♦Feb 15 '13 at 15:41

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

• Try v=u+at for acceleration. – aries0152 Oct 22 '12 at 12:16
• But I don't know how long it took the bike to cover the stretch. I only know Vi (0), Vf (190km/h), and d (402m). – pilau Oct 22 '12 at 12:33
• You are correct that you need more than just that equation. The other kinematic equation you might want to use is $x=1/2 a t^2$, which uses the fact that you started from rest and had a constant acceleration. I'd advise reading a basic kinematics tutorial online. – Alan Rominger Oct 22 '12 at 15:32
• I advise reading ja72's answer below ;) – pilau Oct 22 '12 at 15:47

For constant acceleration you have two equations that you need to solve for time $t$ and acceleration $a$
$$x = \frac{1}{2} a t^2$$ $$v = a t$$
$$\left(402\, {\rm m}\right) = \frac{1}{2} \, a \,t^2$$ $$\left( 190 {\rm \frac{km}{hr}}\right) \left( \frac{1000 {\rm \frac{m}{km}}}{3600 {\rm \frac{sec}{hr}}} \right) = \left(52.777 {\rm \frac{m}{sec}} \right)\, a \,t$$