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

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Homework Help Overview

The discussion revolves around a physics problem involving two hockey players on ice, focusing on calculating the meeting point of a chasing player who accelerates and an opponent who moves at a constant speed. The problem includes aspects of kinematics and motion equations.

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Approaches and Questions Raised

  • Participants discuss the equations of motion for both players, with some attempting to set their positions equal to find the meeting point. Questions arise regarding the correct application of kinematic equations, particularly for the accelerating player versus the one moving at constant speed.

Discussion Status

Some participants have provided guidance on how to incorporate the distance traveled by the opponent during the initial delay before the chase begins. There is an ongoing exploration of the equations needed to solve for the time it takes for the chasing player to catch up.

Contextual Notes

The initial separation distance due to the opponent's speed and the time delay before the chase starts is noted as a critical factor in the problem setup.

Eric [Tsu]
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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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