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

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  • $\begingroup$ Try v=u+at for acceleration. $\endgroup$ – aries0152 Oct 22 '12 at 12:16
  • $\begingroup$ 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). $\endgroup$ – pilau Oct 22 '12 at 12:33
  • $\begingroup$ 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. $\endgroup$ – Alan Rominger Oct 22 '12 at 15:32
  • $\begingroup$ I advise reading ja72's answer below ;) $\endgroup$ – pilau Oct 22 '12 at 15:47

For constant acceleration you have two equations that you need to solve for time $t$ and acceleration $a$

$$ x = \frac{1}{2} a t^2 $$ $$ v = a t $$

It is up to you from here.

$$ \left(402\, {\rm m}\right) = \frac{1}{2} \, a \,t^2 $$ $$ \left( 190 {\rm \frac{km}{hr}}\right) \left( \frac{1000 {\rm \frac{m}{km}}}{3600 {\rm \frac{sec}{hr}}} \right) = \left(52.777 {\rm \frac{m}{sec}} \right)\, a \,t $$


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