# College Physics Problem 2.27

#### In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes $3.33\times 10^{-2}\:s$, calculate the distance over which the puck accelerates.

SOLUTION:

The best equation that can be used to solve this problem is $\Delta{x}=v_{ave}t$. That is,

$\Delta{x}=v_{ave}t$

$\Delta{x}=\left(\frac{8\:m/s+40\:m/s}{2}\right)\left(3.33\times 10^{-2}\:s\right)$

$\Delta{x}=0.7992\:m$

Therefore, the distance over which the puck accelerates is 0.7992 m.

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