Question

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.

Answer 1

0.766 m

Step-by-step 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

`v=u+at`

`40=6+a * 3.33* 10^(-2)`

a = 1021.02 m/s^2

Now to find the distance, use third equation of motion

`v^2=u^2+ 2* a * s`

`40^2=6^2+ 2* 1021.02 * s`

s = 0.766 m

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