Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skates by with the puck. After 2.60 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.50 m/s2, determine each of the following. (a) 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 has he traveled in that time?

Answer 1

1)

Let the time taken by 2nd player be t s

Then 1st player has been moving for t+2.60 s

distance moves by player 1 = v*time

= 4.0*(t+2.60)

= 4.0*t + 10.4

distance moves by player 1 = 0.5*a*t^2

= 0.5*0.50*t^2

=0.25*t^2

Since both distances are equal:

4.0*t + 10.4 = 0.25*t^2

0.25*t^2 - 4.0*t - 10.4 = 0

This is quadratic equation (at^2+bt+c=0)

a = 0.25

b = -4

c = -10.4

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 26.4

roots are :

t = 18.28 and t = -2.276

since t can't be negative, the possible value of t is

t = 18.3 s

Answer: 18.3 s

2)

d = 0.25*t^2

= 0.25*18.3^2

= 83.7 m

Answer: 83.7 m