Finding Time and Vf: Car chase problem

  • #1
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Here's the problem I was given:
Kyle is driving at a constant 45.0 m/s when he passes his street racer friend, Cameron. After a 4.00 second delay to get the car started and into gear, Cameron starts chasing Kyle with a constant acceleration of 2.00 m/s/s. How far will Cameron have to drive to catch Kyle, what will be Cameron's top
velocity, and how long will Cameron drive?

Somehow we are supposed to use the quadratic formula.
We haven't done that in class with this type of problem, only parabolic problems, and I don't know how to correlate the two. (ax^2+bx+c) I don't have enough information to use an equation like Vf=Vi+at or Xf=1/2at^2+Vi+Xi either.

What I've tried so far:

With the three graphs given, I could find the small bit of rectangular area in the first 4 seconds, but that doesn't really help me find anything. I tried different ways of attempting to find the area of the triangle that's formed, but I'm getting answers that don't make sense.

Can someone please explain what I need to do in a step-by-step format? I'm not sure how to start this.
Thank you!!
 
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  • #2
haruspex
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I don't have enough information to use an equation like Vf=Vi+at or Xf=1/2at^2+Vi+Xi
You do.
If Kyle passes Cameron at time 0, what is Kyle's position at time t? What is Cameron's position at time t (for t > 4s)?
 
  • #3
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You do.
If Kyle passes Cameron at time 0, what is Kyle's position at time t? What is Cameron's position at time t (for t > 4s)?
Kyle would be 45m further than Cameron for each 1 of the 4 seconds (or a total of 180m further) Cameron at t>4s would increase by 2.0m/s/s...
 
  • #4
haruspex
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Kyle would be 45m further than Cameron for each 1 of the 4 seconds (or a total of 180m further) Cameron at t>4s would increase by 2.0m/s/s...
No, I mean using the equations you quoted and said you had not enough information to use.

(This is weird. I replied on this thread 12 hours ago but my reply disappeared so I had to type it in again.)
 

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