Approaching a Cops and Robbers Chase Problem Using Equations of Motion

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SUMMARY

The discussion focuses on solving a cops and robbers chase problem using equations of motion. The scenario involves a robber's van with a 1 km head start, accelerating at 0.50 m/s², while the police car accelerates at 0.60 m/s². The key equations utilized include the classic motion equation s = (u * t) + (½ * a * t²) to determine when the speeds of both vehicles will be equal and when the police will catch the robbers. The solution involves setting the distances equal, accounting for the head start, and solving for time.

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  • Understanding of kinematics and equations of motion
  • Familiarity with acceleration concepts in physics
  • Ability to solve quadratic equations
  • Knowledge of basic algebraic manipulation
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  • Learn how to solve problems involving relative motion
  • Explore real-world applications of kinematics in law enforcement scenarios
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Students studying physics, educators teaching kinematics, and anyone interested in applying mathematical concepts to real-world scenarios involving motion.

Deep Dhillon
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Homework Statement



A cops and robbers chase is set up as follows

1. Both the police car and the robber's can start from rest
2. The robbers have 1 km head start
3. The robber's van accelerate at 0.50 m/s
4. The police car accelerate at 0.60 m/s

Use famous 5 equations to answer the following
A. When will the speed of the cop car and the robber van be the same
B. When will the cops catch the robbers ?[/B]

Can someone please give an idea how to approach this answer. This would be very helpful if someone tell m ethe answer!
 
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You know that when they meet, the distance covered by the robber = the distance covered by the cops + 1000 metres, and the time at this point is equal for both.
You could say that :
( Based on the classic equation : s = ( u * t ) + ( ½ * a * t² ) )
½ * 0.6 * t² = ( ½ * 0.5 * t² ) + 1000
 
dean barry said:
You know that when they meet, the distance covered by the robber = the distance covered by the cops + 1000 metres, and the time at this point is equal for both.
You could say that :
( Based on the classic equation : s = ( u * t ) + ( ½ * a * t² ) )
½ * 0.6 * t² = ( ½ * 0.5 * t² ) + 1000
Dean, yet again you give too much help in one go. It would have been fine without that last line. Please stop doing this.
 

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