Rugby Player Collision Question-

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SUMMARY

The discussion focuses on a physics problem involving two rugby players colliding, with one player accelerating from rest at 0.5 m/s² and the other maintaining a speed of 3.1 m/s. The key equations of linear kinematics are employed to determine the time until collision, the final speed of the accelerating player, and the distance each player has run. The solution involves isolating displacement and utilizing the quadratic formula to find the time of collision, confirming that both players meet simultaneously.

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Rugby Player Collision Question--

Homework Statement



Two rugy players are running towards each other. They are 37 m apart. If one is accelerating from rest at 0.5m/s² and the other was already moving at 3.1 m/s and maintains her speed,

a) how long before they crunch together?
b) how fast was the accelerating player going?
c) how far has each player run?

Homework Equations



In this situation: The five equations of linear kinematics

v2 = v1 + a∆t
∆d = ½(v2 + v1)∆t
∆d = v1∆t + ½a∆t²
∆d = v2∆t - ½a∆t²
v2² = v1² + 2a∆d

The Attempt at a Solution



37-d= a(d/v)²/2
37-d= a(d²/v²)/2
74-2d= a(d²/v²)...

And I trailed off from there as I was confused. Now I know you may have to isolate the displacement in this case, however I have not yet encountered a question like this before and it was not clearly explained to me beforehand. Hopefully I'll know what to do through these forums.
 
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One travels d and the other travels 37-d.

They meet at the same time, so find two equations in the two unknowns d and t.
 
I thought it revolved around that, thank you. I used the quadratic formula and found two roots, which was my time for both rugby players, and it went smoothly from there.
 

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