n a slap shot, a hockey player accelerates the puck from a velocity of 5 m/s to 30 m/s in the same direction. If the puck moves over a distance of 10 m during this process, what was the acceleration?

Given data
*The given initial velocity of the puck is u = 5 m/s
*The given final velocity of the puck is v = 30 m/s
*The given distance is s = 10 m
The formula for the acceleration 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{(30)^2-(5)^2}{2\times10} \\ =43.75m/s^2 \end{gathered}[/tex]Hence, the acceleration of the puck is a = 43.75 m/s^2