It's impossible to give a precise answer to this, given all the unknowns. But it's possible to make some estimates:
1. If it took 33 seconds for the cop to start the chase, assuming the speeder going at a steady 86 MPH he would have a head start of 0.78 miles. Of course it takes time for the cop's car to get up to 86 MPH (or more), so the speeder's lead may have been greater than that.
2. Given that the chase on road B lasted 426 seconds and covered 8.2 miles, the cop was traveling at an average speed of 69.3 MPH. We know he started the section at 30 MPH, and slowed to 65 MPH for the corners, so his top speed was clearly greater than that. How much greater is impossible to say from the data provided.
3. It would be really helpful to know whether the cop was able to make up any ground during the chase on Road A. If we assume that the speeder's lead was still 0.78 miles when he turned onto Road B, then he travelled 8.2 - 0.78 = 7.4 miles in 426 seconds, for an average speed of 62.6 MPH on Road B. In other words his average speed on Road B was 7 MPH slower than the cop's. However, if the cop made up some ground while on Road A, so that the speeder's lead was less than 0.78 miles on Road B, then the speeders's average speed is higher than 62.6 MPH (though less than the cop's average speed). On the other hand if his lead was greater than 0.78 miles at the turn then his average speed on Road B would be less than 62.9 MPH. Again, top speed is impossible to determine from the data given.
Hope this helps, but I would love to hear the back story on this question!
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Last edited by ChipB; Jul 12th 2013 at 07:08 AM.
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