Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 13 m/s, skates by with the puck. After 1.5 s, the first player makes up his mind to chase his opponent. (a) If he accelerates uniformly at 4.0 m/s2, how long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.) (b) How far as he traveled

Answer 1

Let us say that:

1 = 1st player notation

2 = 2nd player notation (the opponent)

 

a. First let us establish the distance travelled by the 2nd player:

d2 = 13 m/s * (t + 1.5)

d2 = 13 t + 19.5

 

Then the distance of the 1st player:

d1 = v0 t + 0.5 a t^2         (v0 initial velocity = 0 since he started from rest)

d1 = 0.5 * 4 m/s^2 * t^2

d1 = 2 t^2

 

The two distances must be equal, d1 = d2:

2 t^2 = 13 t + 19.5

t^2 – 6.5 t = 9.75

Completing the square:

(t – 3.25)^2 = 9.75 + (- 3.25)^2

t – 3.25 = ±4.5

t = -1.25, 7.75

Since time cannot be negative, therefore:

t = 7.75 seconds

 

So he catches his opponent after 7.75 seconds.

 

b. Using the equation:

d1 = 2 t^2

d1 = 2 * (7.75)^2

d1 = 120.125 m

 

So he travelled about 120.125 meters when he catches up to his opponent.