A hockey player is standing on his skates on a frozen pond when an opposing play

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 3.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 3.0 m/s2, how long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)

I know the distances must be equal so i have

D(opponent) = D(player)
D = rt = D = (Vi)(t) + 1/2(a)(t^2)

13t + 39 = 1/2(3)(t^2)

set up quadratic equation:

1.5t^2 - 13t - 39 = 0

and put into quadratic formula

a = 1.5
b = -13
c = -39

-(-13) ± (-13)^2 -4(1.5)(-39)
2(1.5)

but i have a negative under the radical..... WHERE HAVE I GONE WRONG!!!!????? i have literally done this problem a million times..... need help desperately........

Explanation / Answer

-(-13) ± v(-13)^2 -4(1.5)(-39)/2(1.5) -(-13) ± v(169+4*1.5*39) /3 = (13 ± 20.07)/3 = 33.07/3 or -7.07/3 neglect negative value, = 33.07/3 = 11.02 hope this helps.

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