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 thisshot takes $3.33 \times 10^{-2} \textrm{ s}$, calculate the distance over which the puck accelerates.

$0.799 \textrm{ m}$

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This is College Physics Answers with Shaun Dychko. The hockey puck begins at a velocity of 8.00 meters per second and has a final velocity of 40 meters per second and the slap shot takes 3.33 times 10 to the minus 2 seconds to impart the acceleration which causes the speed to increase. So, we want to find out over what distance did the hockey stick contact the hockey puck, causing it to accelerate. And we know that displacement is initial position plus average velocity times time. And average velocity is the final velocity plus initial velocity divided by 2. And, this is true when you have constant acceleration which we assume we do here. And then, we plug in numbers. So, we have initial position of 0 plus a final velocity of 40 meters per second plus initial of 8 meters per second divided by 2, times by 3.33 times 10 to the minus 2 seconds. And that gives .799 meters or an answer of about 80 centimeters. It's the distance over which the hockey stick is accelerating the puck.