Question
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.
Step 1
The average velocity is given by the sum of the initial and final velocities divided by two. In mathematical terms, this is represented as: \[v_{avg} = \frac{{v_{f} + v_{i}}}{2}\] where \(v_{f}\) is the final velocity and \(v_{i}\) is the initial velocity. Show more…
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During a slap shot, a hockey player accelerates the puck from a velocity of $8.00 \mathrm{m} / \mathrm{s}$ to $40.0 \mathrm{m} / \mathrm{s}$ in the same direction. If this shot takes $3.33 \times 10^{-2} \mathrm{s}$, what is the distance over which the puck accelerates?
In a slap shot, a hockey player accelerates the puck from a velocity of 11.00 m/s to 60.0 m/s in the same direction. If this takes 3.33 Ă— 10^(-2) s, calculate the distance over which the acceleration acts. _____________=m
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