PHYSICS Question?

Question

A hockey player takes a slap shot at a puck at rest on the ice. The time of contact is 0.0146 s. The puck glides over the ice for 12.8 ft w/o friction, at which point it runs over a concrete surface. The puck then decelerates at 11.1 ft/s^2. If the velocity of the puck is 38 ft/s after traveling 84.6 ft, what is the average acceleration of the puck.

Part 2. How far does the puck travel before stopping.

Part 3. What is the total time the puck is in motion, neglecting contact time?

Answer 1

Step 1: Use Torricelli's equation to find the initial velocity:
vf^2 = vi^2 + 2ad
-vi^2 = -vf^2 + 2ad
vi^2 = (38 ft/sec)^2 - 2 * -11.1 ft/sec^2 * (84.6 - 12.8)
Note: a is negative because it's decelerating)
vi^2 = 1444 + 1593.96 = 3037.96
vi = 55.118 ft/sec

To find the time add the distance over the ice / initial velocity + the change in velocity over acceleration while on concrete:
t = d/vi + (vf-vi)/a
t = 12.8 / 55.118 + 17.118/11.1
t = 0.2322 + 1.5422
t = 1.7744 sec

The average acceleration is:
ice acceleration * ice time / total time + concrete acceleration * concrete time / total time
0 * .2322 / 1.7744 + 11.1 * 1.5432 / 1.7744
avg acceleration = 9.65 ft/sec ^2 (solution #1)

2.) Again, use Toricelli's equation:
vf^2 = vi^2 + 2ad
0 = (55.118 ft/sec)^2 + 2 * -11.1 * d
3037.994 = 22.2d
d = 137 ft (solution)

3.) Total time = time over ice + time over concrete.
time over ice was 0.2322 sec.
Time over concrete = v/a = 55.118 / 11.1 = 4.966 sec
Total time = 5.20 sec (solution)