Kinematics: Motion in One Direction: Car Chase

AI Thread Summary
A car traveling at a constant speed of 24.0 m/s passes a trooper who starts a chase with a constant acceleration of 3.0 m/s² one second later. To determine when the trooper overtakes the car, two distance equations must be set up: one for the car and one for the trooper, accounting for the trooper's delayed start. The discussion emphasizes using relative velocity or a single equation to simplify the problem. Participants suggest drawing velocity-time graphs and equating the distance equations to find the time at which the trooper catches up. The original poster has not responded to the guidance provided.
DracoMalfoy
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Homework Statement


A car traveling at a constant speed of 24.0m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets off in a chase with a constant acceleration of 3.0m/s^2.

A) How long does it take the trooper to overtake the speeding car?

B) How fast is the trooper going at the time?

Homework Equations


  • Vf = Vi+a(t)
  • Δd = Vi(t)+1/2(a)(t)^2
  • Vf^2 = Vi^2+2(a)(Δd)
  • Average Velocity: Δd/Δt
  • Average Acceleration: Vf-Vi/Tf-Ti
Vf: Final Velocity
a: Acceleration
t: Time
Vi: Initial Velocity
Δd: Displacement

The Attempt at a Solution


[/B]
I tried to draw this out first. I know that the Initial Velocity for the trooper has to be 0m/s since they stay hidden behind a billboard before the car passes. and the Acceleration is 3.00m/s^2. I think that a is asking for the time... and I'm guessing that the second question is asking for Final Velocity...

Trooper

Vi: 0m/s
a: 3.00m/s^2
t: ?
Vf: ?

But is 24.0m/s the final velocity for the Car? Or the Initial? Or both? And do I look for the final velocity of the trooper first?
 
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The car is at constant speed for the whole time. When are their positions equal?
 
Cutter Ketch said:
The car is at constant speed for the whole time. When are their positions equal?

Im not sure where to start to solve this. Do they each require a separate equation?
 
Cutter Ketch said:
When are their positions equal?
Hint, hint
 
DracoMalfoy said:
Im not sure where to start to solve this. Do they each require a separate equation?
Yes exactly, write down one equation of distance as function of time for the speeding car, and another equation for the trooper. Take as t=0 the moment the speeding car passes the billboard. Make notice that the trooper starts the chase after 1sec has elapsed in order to make the correct equation of distance for the trooper.
Once you make the two equations equate the right hand sides of the two equations to get a third equation which will have one unknown the time t at which the trooper reaches the speeding car. You got to solve that 3rd equation.
 
DracoMalfoy said:
Im not sure where to start to solve this. Do they each require a separate equation?

Draw velocity time graphs for the two vehicles - that is the best place to start.
 
DracoMalfoy said:
Im not sure where to start to solve this. Do they each require a separate equation?

Instead of two separate equations you could use one single equation for the distance between the two cars.
 
Ray Vickson said:
Instead of two separate equations you could use one single equation for the distance between the two cars.
This single equation can be made by subtracting the two separate equations of distance. But it can be made in a more straightforward manner by using the concept of relative velocity, is that what you had in mind?
 
Delta² said:
This single equation can be made by subtracting the two separate equations of distance. But it can be made in a more straightforward manner by using the concept of relative velocity, is that what you had in mind?

I am leaving all the details for the OP to supply. Hints are the most I am willing to give.
 
  • #10
Ray Vickson said:
I am leaving all the details for the OP to supply. Hints are the most I am willing to give.

You are all talking to yourselves. Delta told him exactly what to do in post 5, and the OP hasn’t been heard from since.
 
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