How long does it take for the first hockey player to catch his opponent?

  • Thread starter willingtolearn
  • Start date
In summary, the problem involves a hockey player chasing an opposing player who is moving at a constant speed of 12 m/s. After 3 seconds, the first player decides to chase and accelerates at 4 m/s^2. The goal is to determine how long it will take for the first player to catch up to the opposing player assuming they remain at constant speed. By equating the displacements of both players, the time can be solved for and determined.
  • #1
willingtolearn
62
0

Homework Statement


A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with they puck. After 3.0s, the first player makes up his mind to chase his opponent. 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)

---------------------------------
The given:
a hockey player that is moving:
speed = 12 m/s
t = (3 + t)

a first hockey player:
vo = 0
a = 4 m/s2

First i tried to find the distance after 3 sec that a moving hocker player travelled.
d = 36 m, so after 3 sec he traveled 36 m. Then i tried to find the remain distance, which i stuck at this points. Can someone help ?
 
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  • #2
I dealt with a similar problem recently, one thing i did was find out what variables they shared, for example if they catch up the Distance is the same. Try making equations for what you have that relate both players and then subsitute solving for t.
 
  • #3
we don't know t and d, so we have 2 variable, how can i solve them
 
  • #4
willingtolearn said:

Homework Statement


A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with they puck. After 3.0s, the first player makes up his mind to chase his opponent. 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)

---------------------------------
The given:
a hockey player that is moving:
speed = 12 m/s
t = (3 + t)

a first hockey player:
vo = 0
a = 4 m/s2

Yes, this is exactly right. Now you just have to equate displacements. What is the displacement of the first hockey player... what is the displacement of the second... set them equal, solve for t, and you get your time.

The first hockey player has been moving for t+3 seconds... the second one for t seconds.
 
  • #5
Is this right (1/2)*4t^2=(3+t)12
 
  • #6
willingtolearn said:
Is this right (1/2)*4t^2=(3+t)12

Yup. Looks good to me.
 
  • #7
Thanks learningphysics !
 

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