Related Rates of Planes: Calculating Distance Change at Noon

shanshan
Messages
23
Reaction score
0

Homework Statement


At ten am, a plane traveling east at 100km/hr is 500 km west of a jet traveling south at 200km/hr. At what rate is the distance between them changing at noon?


Homework Equations





The Attempt at a Solution


2aa' + 2bb' = 2cc'
2(300)(-100) + 2(400)(200) = (1000)c'
-60000 + 160000 = 1000c'
c' = 1000 km/h

i need 186.1 km/h though.
 
Physics news on Phys.org
At noon, I find that the planes are moving apart at a rate of 100 km/hr at noon. I checked a couple of other times to see the rate at which the planes are moving apart. At 11am, the rate of change of distance was 0 km/hr. at 1pm, the rate of change of distance was about 158.1 km/hr. At 2pm, the rate of change is about 186.1 km/hr.

Are you sure you've given us the right times?

BTW, these are a couple of very slow planes. Many planes stall at speeds of 100km/hr, and a speed of 200km/hr for a jet is likewise very slow.
 
yes, these are the times listed on my sheet. there may be a typo though.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top