Calculating Distance Traveled: Police Chase Scenario

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SUMMARY

The discussion centers on calculating the distance traveled by a police trooper chasing a speeding car. The car moves at a constant speed of 129 km/hr, equivalent to 35.83 m/s, while the trooper accelerates at 2.7 m/s². Using the equation d = v(initial)t + 0.5at², participants seek to determine the time and distance at which the trooper overtakes the car. The key challenge is to establish the positions of both the trooper and the car as functions of time and find the intersection point.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = v(initial)t + 0.5at²
  • Knowledge of unit conversion between kilometers per hour and meters per second
  • Familiarity with concepts of constant speed and constant acceleration
  • Ability to solve quadratic equations
NEXT STEPS
  • Learn how to convert speed units from km/hr to m/s for accurate calculations
  • Study the derivation and application of kinematic equations in physics
  • Explore methods for solving quadratic equations to find time of intersection
  • Investigate real-world applications of acceleration and pursuit scenarios in physics
USEFUL FOR

Students in physics courses, educators teaching kinematics, and anyone interested in practical applications of motion equations in real-life scenarios.

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Homework Statement



A car traveling at a constant speed of 129 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 2.7 m/s2. How far does the trooper travel before he overtakes the speeding car?



Homework Equations



d= v(initial)t + .5at^2


The Attempt at a Solution



d= 35.83m/s(t) + .5(2.7)(t)^2 + 35.83

I don't understand what needs done, thanks
 
Physics news on Phys.org
What is the position of the trooper at time t?

What's the position of his target at time t?

When are they in the same place?
 

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