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 :

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

[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]

Substitute the known values in the above expression as

[tex]\begin{gathered} a=\frac{(32.6)^2-(0)^2}{2\times128} \\ =4.15m/s^2 \end{gathered}[/tex]

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