Question

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?

Answer 1

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

`\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=((30)^2-(5)^2)/(2*10) \\ =43.75m/s^2 \end{gathered}`

Hence, the acceleration of the puck is a = 43.75 m/s^2