Two football players run at each other

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SUMMARY

Two rugby players, A and B, are running towards each other from a distance of 37 meters. Player A accelerates from rest at 0.5 m/s², while Player B maintains a constant speed of 3.1 m/s. To determine the time until they collide, the distance each player runs, and the final speed of Player A, simultaneous equations derived from the second equation of motion are utilized. The key equations involved are s = ut + 1/2 * at² for both players, leading to a solution for the time (t) and distance (x).

PREREQUISITES
  • Understanding of kinematics, specifically the equations of motion.
  • Familiarity with acceleration and uniform velocity concepts.
  • Ability to solve simultaneous equations.
  • Basic knowledge of physics terminology related to motion.
NEXT STEPS
  • Study the second equation of motion in-depth: s = ut + 1/2 * at².
  • Learn how to solve simultaneous equations effectively.
  • Explore practical applications of kinematics in sports physics.
  • Investigate the effects of acceleration on collision outcomes in physics simulations.
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Students studying physics, particularly those focusing on kinematics, as well as educators seeking examples of motion problems in sports contexts.

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Homework Statement



Two rugby players are running towards each other. They are 37m apart. If one is accelerating from rest at 0.5m/s^2 and the other was already moving at 3.1m/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



v = at
Vf = Vi + at
Vf^2 = Vi^2 + 2ad

The Attempt at a Solution



Sorry, I didn't learn this and my attempt was only a diagram and some rough equations.
 
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hi,
i tried to solve the problem, but i think that the numbers given are a little incorrect. anyways, here is my procedure:

1)Let us take the two footballers as A and B respectively, so, A is at rest and starts accelerating at 0.5m/s^2, and B moves with an uniform velocity of 3.1m/s.

2)at some point of time the two are going to crash, so take that point to be O. the distance from the starting point of A to O is x meters (let us consider). hence, the distance from the starting point of B to O will be 37 - x meters. sorry for the rough diagram.

A ---- x meters ----- O ------------ 37-x meters ----------------B

3)now, use the 2nd equation of motion to solve, s = ut + 1/2 * at^2, where u is initial velocity, and s is the distance. here the time t will remain constant because both will crash at the same time.

for A,

s = ut + 1/2 * at^2
=> x = 1/2 * 0.5t^2 ----------- (1)

for B,

s = ut + 1/2 * at^2
=> 37 - x = 3.1t ------------- (2)

you can now solve (1) and (2) to get answers to a) and c) by solving simultaneous equations for the values for t and x respectively.

4)after you get all the previous values, getting an answer for b) becomes very easy with numerous options for equations. (eg: you could use v = u + at, where v is final velocity)
 

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