In a slap shot a hockey player accelerates the puck from a velocity of 5.00 m/s

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In a slap shot a hockey player accelerates the puck from a velocity of 5.00 m/s to 60.0 m/s in the same direction. If this takes 3.33 ✕ 10-2 s, calculate the distance over which the acceleration acts.

 m

Sep 8th, 2015

Thank you for the opportunity to help you with your question!

Average velocity= displacement /time

velocity= 60m/s – 5m/s= 55m/s

time = 3.33x10-2 s

distance=?

Distance= ∆velocity x time

=55 x 3.33x10-2

=1.8315m


Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 8th, 2015

Sry,but the answer was wrong

Sep 8th, 2015

i wil correct it please

Sep 8th, 2015

acceleration, a, is: a = (change in velocity)/change in time. 
change in velocity = (60 - 5)m/w = 55 m/s 
change in time = 3.33(10^-2) 
a = 55/(3.33)(10^-2) = 16.5(10^2) m/s^2 
distance = (1/2)at^2 = (1/2)(16.5)(10^2)(3.33^2)((10^-2)^2) = 91.48 x 10^-2 m = 91.48 cm.

Sep 8th, 2015

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