DJBruce Posted July 8, 2008 Posted July 8, 2008 At this Camp I am attending a Professor of Statistics posed this question to the class: If I want to go from point A to point B and back to point A (so a round trip) and the distance from A to B was 30 miles and I wanted to have an average speed of 60 mph how fast would I have to drive if it took me one hour to go from A to B. So basically you want to find out how fast you need to go from B to A to average a speed of 60mph. If you know the answer could you please explain it. I think I know it but am not sure.
ecoli Posted July 8, 2008 Posted July 8, 2008 answer hidden below [hide]90 mph it's pretty simple. The way there is 30mph and average velocity is 60 mph. Ave Velocity = (V1 + v2) / 2 plug in and solve for V2 [/hide]
Aeternus Posted July 8, 2008 Posted July 8, 2008 answer hidden below [hide]90 mph it's pretty simple. The way there is 30mph and average velocity is 60 mph. Ave Velocity = (V1 + v2) / 2 plug in and solve for V2 [/hide] I don't think that's correct, you are averaging over distance, not time. I believe you can't do it. Take the speed we travel back with to be [math]x[/math]. We have [math] \textrm{average speed} = \frac{\textrm{total distance}}{\textrm{total time}} [/math]. We have the total distance as 60, and the time is going to be the 1 hour we have already used, added to the time it will take us to cover the remaining 30 miles, so [math]\frac{30}{x}[/math]. This gives us [math] 60 = \frac{60}{1 + \frac{30}{x}} [/math] [math]1 + \frac{30}{x} = 1 [/math] [math]\frac{30}{x} = 0 [/math] As you can see, there is no solution. If you could travel for another hour at 90 mph, then it would be fine, but you only have 30 miles left to cover.
insane_alien Posted July 8, 2008 Posted July 8, 2008 it is indeed impossible. to have an average speed of 60 mph he must complete the journey within 1 hour. to do this he needs to travel infinitely fast back to point a. as this is impossible(in the real world and a hypothetical one where the laws ofphysics must be obeyed) he cannot do it.
YT2095 Posted July 8, 2008 Posted July 8, 2008 since it only said A to B and didn`t specify the Route or Quantity, might it help you Math guys out if you consider A to B to A to B (doing it twice), giving the net effect of just A to B? OR traveling in a Zig-Zag or Arc? just a thought
DJBruce Posted July 9, 2008 Author Posted July 9, 2008 Ecoli that is exactly what I thought first but he said it was wrong and has now revealed it was impossible. Thanks for the help in understanding why seeing the math helped a lot.
Kyrisch Posted July 9, 2008 Posted July 9, 2008 If I want to go from point A to point B and back to point A (so a round trip) and the distance from A to B was 30 miles and I wanted to have an average speed of 60 mph how fast would I have to drive if it took me one hour to go from A to B. This is not solvable. Too much information is given... The question is not self-consistent.
D H Posted July 9, 2008 Posted July 9, 2008 (edited) There is not too much information. For example, a solution would exist if it took 45 minutes to go from A to B rather than 60 minutes. The information given does not yield a solution. Insane Alien got the answer in post #4. DJBruce, think of it this way: If you want to have an average speed of 60 mph or more, you will need to make the round trip in one hour or less. If you start at noon you have to be back to the starting point by 1 PM. You have a deadline. If the deadline passes and you still on the road, game over. It took an hour to go from A to B, so the deadline passed when you reached point B. Game over. Edited July 9, 2008 by D H
YT2095 Posted July 9, 2008 Posted July 9, 2008 what happens if he drives for 15 miles from point A and realises he`s forgotten something so drives back to A and then sets off again, total time on the road is 1 hour if he drove at 60mph. is that allowed?
john5746 Posted July 9, 2008 Posted July 9, 2008 I think everyone is correct in their own way. It does not state in the problem that the total trip time must be one hour and that the trip back must be 30 miles. If you assume the trip back must be 30 miles, then you have run out of time(or miles). If you can take a scenic route as YT suggests, then traveling 90 miles back to A will give the 90mph answer ecoli provided.
D H Posted July 9, 2008 Posted July 9, 2008 If you can take a scenic route as YT suggests, then traveling 90 miles back to A will give the 90mph answer ecoli provided. Driving 90 mph from B to A, back to B, and then back to A makes the average speed 60 mph over two round trips, not one round trip. Besides, this is not what ecoli proposed as a solution. Ecoli made what is a very typical mistake in calculating the average speed. The average speed on a trip is not the arithmetic mean of the speeds attained on individual legs of the trip. The point of this exercise is to highlight that common mistake.
john5746 Posted July 9, 2008 Posted July 9, 2008 Driving 90 mph from B to A, back to B, and then back to A makes the average speed 60 mph over two round trips, not one round trip. I didn't say that. Picture a 30 mile crappy dirt road going from A to B and also a nice highway bypass that is 90 miles in length between the two. Besides, this is not what ecoli proposed as a solution. Ecoli made what is a very typical mistake in calculating the average speed. The average speed on a trip is not the arithmetic mean of the speeds attained on individual legs of the trip. The point of this exercise is to highlight that common mistake. cmon, where's the love? can't we all get a little partial credit? I would suck as a teacher!
ecoli Posted July 9, 2008 Posted July 9, 2008 (edited) Besides, this is not what ecoli proposed as a solution. Ecoli made what is a very typical mistake in calculating the average speed. The average speed on a trip is not the arithmetic mean of the speeds attained on individual legs of the trip. The point of this exercise is to highlight that common mistake. But if you're considering the average speed of the displacement, and not the distance, then the answer would be correct. The original question is too vague, I think, to rule this out. Aeternus says that Average speed = total distance/ total time, but Average velocity is total displacement/total time edit: actually, now that I look again, the OP does ask for ave. speed, so I guess I stand corrected. Edited July 9, 2008 by ecoli
Mag Posted July 10, 2008 Posted July 10, 2008 The question says it will take an hour to go from A to B, but it does not say how fast it will take to go from B to A. The professor just needs to step up his game on the way back to A, with total disregard for the speed limit, and go 90mph so that he may have his average of 60.
ecoli Posted July 10, 2008 Posted July 10, 2008 The question says it will take an hour to go from A to B, but it does not say how fast it will take to go from B to A. The professor just needs to step up his game on the way back to A, with total disregard for the speed limit, and go 90mph so that he may have his average of 60. As we covered already, however, that doesn't work, because the return trip is only 30 miles, if you phrase the answer in this way. Average speed = Total Distance/ Total time. So, if the total distance is 60mi, the only way to have an average of 60mph round trip, is if the total time is one hour: 60 mph = 60miles/ 1 hour However, as the question stated, the first half of the trip took an hour, so obviously, the denominator of this equation is going to be greater than 1. 60 mph = (X miles) / (1 + y hours) You could still get an Ave. speed of 60mph, but only if the distance is greater than 60 miles, since the total time is greater than an hour. If the problem asked for the velocity, than you could answer in terms of displacement, and the question is solvable. However, this is not the case with 'speed.'
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