Question

A motorcycle is uniformly accelerated over a distance of 128 meters. If the original speed of the motorcycle is 0m/s and the final velocity is 32.6m/s, what acceleration did the bike undergo?

Answer 1

Given data

*The given distance is s = 128 m

*The initial speed of the motorcycle is u = 0 m/s

*The final speed of the motorcycle is v = 32.6 m/s

The expression for the acceleration of the bike is given by the kinematic equation of motion as

`\begin{gathered} v^2=u^2+2as \\ a=(v^2-u^2)/(2s) \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} a=((32.6)^2-(0)^2)/(2*128) \\ =4.15m/s^2 \end{gathered}`

Hence, the acceleration of the bike undergoes is a = 4.15 m/s^2