Ahockey 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

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.

answer would be d.) i hope this

explanation:

the answer is a 3.8cm

explanation: