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dumperdery
Joined: 23 May 2010 Posts: 4 Location: Canada

Posted: Mon May 24, 2010 5:04 pm Post subject: Physics Problem 2  Distance, Time Problem 


A hockey player is standing on his skates on a frozen pond when an opposing
player skates by with the puck, moving with a constant speed of 12 m/s. After
3.0 s, the first player makes up his mind to chase his opponent and starts
accelerating uniformly at 3.8 m/s2.
a. How long does it take him to catch his opponent?
b. How far does he travel before he catches up with his opponent? 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Mon May 24, 2010 10:52 pm Post subject: 


This problem can be solved graphically or by writing equations of motion for the distance and time.
For the hockey player moving at constant speed, his equation of motion would be:
s(t) = 12 t
i.e. every second he moves 12 metres.
For the second hockey player, we need the equation of motion that gives the distance covered in terms of the acceleration as a function of time.
s(t) = u t + 0.5 a t^2
Since the second player does not move until 3 seconds after we need replace the time by t3.
Then to find the time you need to equate the two equations and solve for t.
When you know t, you can substitute it back into either of the equations of motion to find the distance.
I make the time, 11.53 sec and the distance 138.44 metres. 

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dumperdery
Joined: 23 May 2010 Posts: 4 Location: Canada

Posted: Tue May 25, 2010 8:49 pm Post subject: 


I got 8.53 seconds and 138 meters.
my equation was
(0.5)(3.(T^2) = 12 (T + 3)
I solved for T using the quadratic formula and got 8.53 seconds. When you plug this into D = (Vi)(T) + (0.5)aT^2 you get 138 meters 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Tue May 25, 2010 10:26 pm Post subject: 


Y, it depend on when the time starts. 11.53 if the clock starts when the first hockey player starts to move. 8.53 seconds, if clock starts when the second player moves. Since the question says how long does it take for him to catch up the answer would be 8.53 secs. 

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