Percent Difference problem

J

jschwa1

New member
Joined
Aug 2, 2010
Messages
15
A plane flew at 37000 ft and 590 mph from one city to another. A car drove between the same locations. What is the percent difference in their travel distances?

I know the formula for percent difference and that I probably need to somehow use d = rt, but it doesn't seem like I have enough information.
 
You are correct. There is not enough information given to calculate anything, really.

We could disregard reality, and assume that the plane and car each moved the same distance because they both started at the same place and went to the same place. That would make the answer trivial.
 
I have the answer to the problem, and it is -.177%, but this still isn't helping me. I guess (ignoring reality) that the plane would have to travel a further distance because it rises to and falls from the 37000 feet. I still don't get it.
 
jschwa1 said:
I guess (ignoring reality) that the plane would have to travel a further distance

I think you mean "(not ignoring reality)".

Such a guess seems reasonable, to me, but then, in making such a guess, we're also assuming that the highway is straight. Who knows? If we had some information about the plane's rate of ascent and rate of descent, perhaps we could do something.

I still don't get it. Neither do I. Not enough information is given to arrive at the posted answer.
Click to expand...
 
I figured it out. Since the plane and the car are actually travelling along an arc, use the formula s = r x theta. The radius of the earth is 20925524.9 feet. Therefore, the percent difference equals (((20925524.9 x theta) - (20925524.9 + 37000)theta)/(20925524.9 x theta) ) x 100 = -.177%
 
You must log in or register to reply here.
Top