I\'ve worked on this for several days-not getting there. A hockey player (1) is

Question

I've worked on this for several days-not getting there. A hockey player (1) is standing on his skates on a frozen pond when an opposing player (2)player 2 and x2, moving with a uniform speed of 12 m/s, skates by with the puck. After 3.0s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2, (a) how long does it take him to catch his opponent? Okay so I know that player 2 has v=12 m/s and his a=0. Player 1 has a=4m/s2 and v=0. I had also manually counted out distance each would be at 1 sec, 2 sec, etc. I first tried variations of a=v-vo/t which didn't work at all. I emailed out tutor (this is an online course-I have NO access to a real teacher), who told me I needed to find an equation for each and set them equal to each other. I found some more help in a physics problem solving book. So I tried x=vt for x2= 1/2at2 for player 1. Thanks!

Explanation / Answer

First, let's find how far the 2nd player gets in the 3 seconds when the 1st player isn't doing anything:
x = vt = 12*3 = 36 m

Now let's find an equation for the distance skated by player 1:
x = 0.5*a*t^2 = 0.5*4*t^2 = 2t^2

Now find an equation for the distance of player 2, taking t=0 when player 1 starts chasing him:
x = x0 + v0t = 36 + 12t

We want to find when the distances are the same. This means that x(player1) = x(player2). Therefore, we can set our two position functions equal to each other:
2t^2 = 36 + 12t

Rewrite this in standard form for quadratic equations, and then apply the quadratic formula to solve for t. You should get:
2t^2-12t-36 = 0
t = -2.196 and t = 8.196

Negative time doesn't make sense, so the answer must be 8.2 seconds (rounding to significant figures).

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