In a slap shot, a hockey player accelerates the puck from a velocity of 6.00 m/s to 40.0 m/s in the same direction. if this takes 3.33 ✕ 10−2 s, calculate the distance over which the puck accelerates.

0.766 m

Explanation:

initial velocity of puck, u = 6 m/s

final velocity of puck, v = 40 m/s

time taken by the puck, t = 3.33 x 10^-2 s

Let the acceleration of puck is a and the distance traveled by the puck is s in the given time.

Use first equation of motion

a = 1021.02 m/s^2

Now to find the distance, use third equation of motion

s = 0.766 m

Thus, the distance traveled by the puck is 0.766 m.

this means that the temperature of both substances will be equal

true is the correct answer

hope i