Can Police Clock Speeding Car?
Date: 06/30/2003 at 10:23:03 From: Michael Subject: Maximum speed of a car traveling over a distance Car A is traveling a 3-mile distance at an average speed of 87 mph. Car B begins the same distance from an initial speed of 0, but accelerates from 0-100 mph in 14 seconds. What must the maximum speed of Car B be in order to catch up to Car A at the end of the 3 miles?
Date: 06/30/2003 at 13:13:09 From: Doctor Jaffee Subject: Re: Maximum speed of a car traveling over a distance Hi Michael, If you were to set up a graph with time on the horizontal axis and rate of speed on the vertical axis, so that rate is a function of time, the graph representing Car A would be a horizontal line, f(t) = 87. I'm going to assume that Car B's velocity increased at a constant rate, so the graph representing Car B (for the first 14 seconds) will be a line segment connecting (0,0) to (14/3600,100). In either case the distance travelled will be the area under the line segment, since d = rt. It will take Car A just a shade over 2 minutes and 4 seconds to travel the three-mile distance. The distance that Car B will travel in the first 14 seconds is the area of the triangle under the segment for the first 14 seconds. Once you have that information, you are nearly finished. Car B now has about 1 minute and 50 seconds to go and you can calculate the distance remaining. Take the distance and divide by the time and you will know the rate required to catch Car A. Give it a try and if you want to check your answer with me or if you have difficulties or other questions, write back to me and I'll try to help you some more. Good luck, - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
Date: 06/30/2003 at 15:55:43 From: Michael Subject: Maximum speed of a car traveling over a distance I have figured that considering acceleration, Car B make the acceleration from 0-100 in 14 seconds in .453 of a mile. I do not believe that to be correct however. Somewhere my math is wrong. I am still having difficulty figuring out the rate of speed for Car B. I figure it should be a parabola, and I am looking for the maximum speed. In the same time, the 14 seconds, I have figured Car A to have traveled .3387 of a mile.
Date: 07/01/2003 at 17:32:18 From: Doctor Jaffee Subject: Re: Maximum speed of a car traveling over a distance Hi Michael, I agree with your calculations that Car A traveled approximately .3387 miles in 14 seconds. The problem is the distance that Car B travels in the same 14 seconds. According to my calculations, Car B travelled exactly 7/36 of a mile, or 0.194444... miles. This calculation is based on my assumption that the acceleration during this interval was constant, which means that the graph of v = f(t) is linear and the distance travelled is the area under the triangle formed. I multiplied (1/2)(14/3600)(100). Your answer is based on the assumption that the acceleration was not constant and that the distance travelled is the area under a parablola. Nowhere in the problem is there a justification for either of our assumptions, but it certainly would be easier if the acceleration were constant. In any case, I would be curious to know what type of acceleration you chose and how you used it to come up with your answer. I hope what I have just written makes sense and provides some clarification for you. Write back and let me know how you got .453 miles and whatever other comments or questions you have. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
Date: 07/01/2003 at 23:47:23 From: Michael Subject: Maximum speed of a car traveling over a distance I don't know exactly how I came up with the calculation of .4 whatever... I got the 14 seconds from a rating the car was given in 0-100. I am trying to beat a speeding ticket, and my reasoning is that I (Car A) could not have been doing 87 mph, as my equipment read between 75-80 over the 3 miles. I want to know the maximum speed that Car B would have had to travel in order to catch me, because he caught me before the 3 miles, but I'm using that for refrence. Any help will be appreciated. Thank you.
Date: 07/02/2003 at 14:54:52 From: Doctor Jaffee Subject: Re: Maximum speed of a car traveling over a distance Hi Michael, I understand the problem a little better now, and it has gotten pretty interesting. It sounds as if what you want to do is argue that if you had been going 87 miles per hour, as the officer contended, and his car was capable of accelerating from 0 to 100 mph in 14 seconds, there is no way that he could have caught up to you within 3 miles after you passed him. But he did catch you in less than 3 miles. Therefore, you must have been going slower than 87 mph. Unfortunately, the numbers don't support your argument according to my computations. If the police car accelerated at a constant rate, the officer would have travelled nearly 1/5 of a mile in 14 seconds and by the time he drove a total of 3 miles could have caught a car travelling 87 mph if the officer drove at a rate of approximately 91 mph. I figure that even if the police car accelerated at a rate whose graph was concave upward, it would only travel about 1/8 mile in the first 14 seconds, but could still catch the other car if the officer maintained a 94 mph speed. The men at the garage where I get my car serviced tell me that, in general, a car that can accelerate from 0 to 100 in 14 seconds travels nearly 1/4 mile. If that were true, the officer would have to maintain only 90 mph after the first 14 seconds to catch you at the end of three miles. Now, you didn't say how much before the end of the three-mile interval the officer caught you. If it were 2.5 miles, that would make a difference. Also, the men at the garage reminded me that if you have changed the tires on your car to a different size from the original tires, that will affect the reading on the speedometer. I hope my comments shed some more light on this discussion. Write back and let me know if what I wrote makes sense to you and what other comments or questions you have. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
Date: 07/03/2003 at 00:20:03 From: Michael Subject: Maximum speed of a car traveling over a distance It looks as if you've been able to answer all my questions. Of course, I haven't taken into account the fact of the traffic behind me, so it could have delayed his exit. He also caught me before the 3 miles. He actually pulled me over at the 3 mile point, and said he was able to clock my speed twice. That's another factor to the problem. I just thought it was a stretch for him to catch me if I had been traveling 27 mph over the speed limit, with traffic, to have caught me so soon. As just as a coincidence, I just replaced a set of tires earlier this week. But I never have been pulled over in order to do a speedometer check. I actually checked later that night, and found that on 3 occasions it took me 49 seconds to travel a mile. I figured that to be approximetly 73 mph. So although my speedometer IS off, it's not off by as much as what he said. But my argument that he would not have been able to catch me is invalid now. Just for curiosity, say he HAD caught me only 2.5 miles from the start, what would the average speed be? Just another variable I guess. Thanks for all your help. - Mike
Date: 07/03/2003 at 09:20:13 From: Doctor Jaffee Subject: Re: Maximum speed of a car traveling over a distance Hi Mike, I'm going to put you to work on this last question. This is, after all, a mathematics education website. You know that if rate of speed is a constant, distance = rate x time. So, assuming that you were travelling 87 mph and your distance was 2.5 minutes, you can easily calculate the time it would take. Now, let's assume that the police officer travelled 1/4 mile in the first fourteen seconds. That would be, of course, under perfect, drag strip conditions, and personally, based on my calculations, I'm skeptical that even then he could cover that much ground. If you figured out how much time it took you to travel the 2.5 miles, subtract 14 seconds from that time to determine how much more time the officer had to catch you. The distance the officer has to travel is now 2.25 miles. Finally, d/t = rate of speed and that it will tell you how fast he needed to go to catch you. Give it a try and if you want to check your solution with me or if you have difficulties or other questions write back to me and I'll try to help you some more. I've enjoyed working on this problem. Good luck, - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
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