Kinematics: Motion in One Direction: Car Chase

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Homework Help Overview

The problem involves kinematics, specifically the motion of a car traveling at a constant speed and a trooper accelerating from rest to catch up. The original poster seeks to determine the time it takes for the trooper to overtake the car and the trooper's speed at that moment.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to establish equations for the positions of both the car and the trooper, considering the trooper's delayed start. Questions arise regarding the initial and final velocities of the car and how to formulate the equations of motion for both vehicles.

Discussion Status

Some participants have suggested writing separate equations for the distance traveled by each vehicle and equating them to find the time at which the trooper catches up. Others have proposed using relative velocity to simplify the problem. There is an ongoing exploration of different approaches without a clear consensus on the best method.

Contextual Notes

There is a mention of the trooper starting the chase one second after the car passes the billboard, which is a crucial aspect of the problem setup. The original poster has not yet provided further input since the initial post.

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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