How Do You Calculate the Meeting Point of Two Hockey Players on Ice?

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To calculate the meeting point of two hockey players, one moving at a constant speed of 15 m/s and the other accelerating at 4.0 m/s² after a delay of 3.5 seconds, the initial distance between them is 52.5 meters. The equations for their positions must be set equal to find the time "t" when they meet. The opponent's position can be described by the equation X = 15t, while the chaser's position is given by X = 52.5 + 0.5(4)(t²). Solving the equation 15t = 52.5 + 2t² leads to a quadratic equation in "t," which can be solved to determine the time taken for the chaser to catch up. This approach effectively combines both players' motion to find the point of intersection on the ice.
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1. A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 15 m/s, skates by with the puck. After 3.5 s, the first player makes up his mind to chase his opponent.

(a) If he accelerates uniformly at 4.0 m/s^2, 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 this time?
I've tried to solve it by coming up with an equation for the position of each player, and set them equal to each other. Equation for the player who is chasing: X = v*t

I'm not sure what formula to manipulate in order to come up with an equation for the position of the player being chased.

Thanks in advance.
 
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My bad, I wrote it opposite. X = v*t is for opponents position
 
Remember the opponent will have traveled some distance in the thinking time as well. Is there any equations in the link I gave that you can see helping you for the chasing player?
 
X = Xi + Vi(t) + 1/2(a)(t^2)

Xi = 52.5
Vi = 0
a = 4

Chaser --> X = 52.5 + 1/2(4)(t^2)
Opponent --> X = v*t

Lost at this point
 
It would be best to add the distance the opponent has traveled onto the opponents equation and just have the acceleration in the chasers equation. From there you want to make them equal and then you will have a quadratic equation in t to solve.
 
Eric [Tsu];1758628 said:
1. A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 15 m/s, skates by with the puck. After 3.5 s, the first player makes up his mind to chase his opponent.

(a) If he accelerates uniformly at 4.0 m/s^2, 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 this time?



I've tried to solve it by coming up with an equation for the position of each player, and set them equal to each other. Equation for the player who is chasing: X = v*t

I'm not sure what formula to manipulate in order to come up with an equation for the position of the player being chased.

Thanks in advance.

The initial separation is (15*3.5)=52.5 m
Let they meet after "t" seconds
15t=0.5*4*(t*t)+52.5
solve for "t"
 
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