Solving 1D Motion: How Fast Must a Police Car Go to Catch a Holdup Car?

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SUMMARY

The problem involves a police car chasing a holdup car that travels at a constant speed of 120 km/h. The holdup car has a 0.50-hour head start, covering a distance of 60 km before the police car begins its pursuit. To catch up in 1.00 hour, the police car must travel a total distance of 180 km, requiring a speed of 180 km/h. This calculation is based on the formula for distance, s = vt, where the total time for the holdup car is 1.5 hours.

PREREQUISITES
  • Understanding of basic kinematics, specifically distance, speed, and time relationships.
  • Familiarity with the formula s = vt for calculating distance.
  • Knowledge of constant velocity motion without acceleration.
  • Ability to solve linear equations involving speed and distance.
NEXT STEPS
  • Review kinematic equations for motion in one dimension.
  • Practice problems involving relative motion and pursuit scenarios.
  • Explore the concept of constant acceleration and its equations.
  • Learn about graphical representations of motion, such as distance-time graphs.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for practical examples of motion problems.

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Homework Statement

A police car leaves in pursuit of a holdup car 0.50 h after the latter has left the scene of the crime at 120 km/h. How fast must the police car go if it is to catch up with the holdup car in 1.00h?


Homework Equations

I'm thinking Vf=Vo+at? i must be wrong, I can't seem to get the probelm started



The Attempt at a Solution

Tried so many but it doesn't seem to come out
 
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I think you can ignore accelaration since no data is given. So it is simply s = vt
How far has the first car gone at constant speed when the police car catches it?
 
then distance = 120(1.5), Which equals 180 km. Thank you for the help I didn't even think of it like that.
 

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