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revisisland24

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Hey guys,
I've been having trouble getting the right equations for this problem.

A police helicopter is flying at 150 mph at a constant altitude of .5 miles above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter, and that this distance is decreasing at 190 mph. Determine the speed of the car.

Try to show the work please because I want to learn how to do this! Thanks!
 
Can you show us what you have tried? This will be far more instructive for you if we can tell you why what you've tried does not work.
 
revisisland24 said:
I've been having trouble getting the right equations for this problem.

A police helicopter is flying at 150 mph at a constant altitude of .5 miles above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter, and that this distance is decreasing at 190 mph. Determine the speed of the car.
Click to expand...

Try to show the work please because I want to learn how to do this! Thanks!
Click to expand...
You have loads of worked examples in your book, your classnotes, and in pages you've (surely) viewed online. So, if you're needing help, obviously one more worked example won't make the difference. Instead, as another poster pointed out, you'll need to show your work (the stuff that led to the "wrong" equations), so we can try to find where you're going astray.

That said, a good place to start would probably be drawing a picture of the situation, showing the right triangle, etc. What triangle did you end up with, and how did you label things? For convenience, let's use "h" for the height of the chopper, "x" for the length along the ground between the chopper and the car, and "R" for the radar-view (direct and "diagonal-ish") distance between the chopper's radar and the oncoming car.

Please be complete. Thank you! ;)
 
Staple gave good advice,

For convenience, let's use "h" for the height of the chopper, "x" for the length along the ground between the chopper and the car, and "R" for the radar-view (direct and "diagonal-ish") distance between the chopper's radar and the oncoming car.

COMMENT:
small assumption: h is vertical, and x is horizontal. That means at right angle to each other. This is like saying the car is not going up hill or down hill.

NEXT:
Take each letter and write what you believe comes after the equal sign:
x =
h=
R=


 
stapel said:
You have loads of worked examples in your book, your classnotes, and in pages you've (surely) viewed online. So, if you're needing help, obviously one more worked example won't make the difference. Instead, as another poster pointed out, you'll need to show your work (the stuff that led to the "wrong" equations), so we can try to find where you're going astray.

That said, a good place to start would probably be drawing a picture of the situation, showing the right triangle, etc. What triangle did you end up with, and how did you label things? For convenience, let's use "h" for the height of the chopper, "x" for the length along the ground between the chopper and the car, and "R" for the radar-view (direct and "diagonal-ish") distance between the chopper's radar and the oncoming car.

Please be complete. Thank you! ;)
Click to expand...

I made a right angle triangle and got 1 mile for the hypotenuse. So, the sides would be .5, sq. root of .75, and 1. Where would I go from there?
 
Bob Brown MSEE said:
Staple gave good advice,

For convenience, let's use "h" for the height of the chopper, "x" for the length along the ground between the chopper and the car, and "R" for the radar-view (direct and "diagonal-ish") distance between the chopper's radar and the oncoming car.

COMMENT:
small assumption: h is vertical, and x is horizontal. That means at right angle to each other. This is like saying the car is not going up hill or down hill.

NEXT:
Take each letter and write what you believe comes after the equal sign:
x =
h=
R=
Click to expand...
I made a right angle triangle and got 1 mile for the hypotenuse. So, the sides would be .5, sq. root of .75, and 1. Where would I go from there?
h=.5 miles, x=sq. root of .75, R = 1 mile
 
The only constant is the helicopter's height above the ground, and so using Bob Brown MSEE's variables, we have via Pythagoras:

\(\displaystyle x^2+\left(\dfrac{1}{2} \right)^2=R^2\)

You are being asked to find \(\displaystyle \dfrac{dx}{dt}\)...any idea how you can find this from the above equation? Will you have everything else you need?
 
MarkFL said:
The only constant is the helicopter's height above the ground, and so using Bob Brown MSEE's variables, we have via Pythagoras:

\(\displaystyle x^2+\left(\dfrac{1}{2} \right)^2=R^2\)

You are being asked to find \(\displaystyle \dfrac{dx}{dt}\)...any idea how you can find this from the above equation? Will you have everything else you need?
Click to expand...

Would that equal dx/dt = R^2-.25/2x?
 
No, R is also a function of time, so it must be differentiated as well whereas the constant is unchanging so its derivative is zero.
 
MarkFL said:
No, R is also a function of time, so it must be differentiated as well whereas the constant is unchanging so its derivative is zero.
Click to expand...

you're right so it would by 2xdx/dt = 2R dR/dt which would equal dx/dt = R/x * dR/dt
 
Yes, now you want to plug in the given data...you are given \(\displaystyle R\) and \(\displaystyle \dfrac{dR}{dt}\), and you can find \(\displaystyle x\)...what should you do with the information regarding the helicopter's speed?
 
MarkFL said:
Yes, now you want to plug in the given data...you are given \(\displaystyle R\) and \(\displaystyle \dfrac{dR}{dt}\), and you can find \(\displaystyle x\)...what should you do with the information regarding the helicopter's speed?
Click to expand...

So, I got 212.43 mph as dx/dt. Would I subtract this speed by the speed of the helicopter?
 
Last edited:
I get a slightly higher value for dx/dt...can you show what you did to get your value? Once we get this ironed out, we'll deal with the helicopter's speed...
 
MarkFL said:
I get a slightly higher value for dx/dt...can you show what you did to get your value? Once we get this ironed out, we'll deal with the helicopter's speed...
Click to expand...

(1)^2 + (.5)^2 = 1.25, which makes R = sq. root of 1.25, dR/dt = -190 mph

-dx/dt = -(sq. root 1.25/1) (-190) = 190 * sq. root of 1.25 which is about 212.43
 
Oh okay I see...you want to let R = 1, not x.
 
Yes, that's the same value I found.

One way to deal with the helicopter's speed is to let \(\displaystyle x_C\) be the car's horizontal coordinate and \(\displaystyle x_H\) be the helicopter's horizontal coordinate, where \(\displaystyle x_H>x_C\) and so:

\(\displaystyle x=x_H-x_C\)

hence:

\(\displaystyle \dfrac{dx}{dt}=\dfrac{dx_H}{dt}-\dfrac{dx_C}{dt}\)

\(\displaystyle \dfrac{dx_C}{dt}=\dfrac{dx_H}{dt}-\dfrac{dx}{dt}\)

we are given: \(\displaystyle \dfrac{dx_H}{dt}=-150\text{ mph}\) and so what do we find?
 
MarkFL said:
Yes, that's the same value I found.

One way to deal with the helicopter's speed is to let \(\displaystyle x_C\) be the car's horizontal coordinate and \(\displaystyle x_H\) be the helicopter's horizontal coordinate, where \(\displaystyle x_H>x_C\) and so:

\(\displaystyle x=x_H-x_C\)

hence:

\(\displaystyle \dfrac{dx}{dt}=\dfrac{dx_H}{dt}-\dfrac{dx_C}{dt}\)

\(\displaystyle \dfrac{dx_C}{dt}=\dfrac{dx_H}{dt}-\dfrac{dx}{dt}\)

we are given: \(\displaystyle \dfrac{dx_H}{dt}=-150\text{ mph}\) and so what do we find?
Click to expand...

We get 219.39 - 150 = 69.4 mph as the car's speed?
 
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