Car Speed: Find the Answer Here

[mathman100]

A plane 2km high is flying at rate of 120 km/h due west and sees an oncoming car. The distance between them is 4km and is decresing at rate of 160 km/h. Find the speed of the car at this moment...
I can't solve this! i guess the 4km is horizontal diatnce or else the plane and car would crash...

 

[Hootenanny]

The fact that you are given the height of the plane suggests that 4km is the actual distance between the object and you will need to use pythag to calculate the linear distance. Show some of your working and I'll see if I can give you some hints.

 

[mathman100]

ok, here is what i thought, i used related rates:
triangle with sides a (height of plane=2), b (sqrt12, found by pythag.theorem) and side c (hypotenuse, =4 km, ditsance from car-plane)
c'=(sqrt12*120)/4
=104
then to find theta
sintheta=2/4, theta = 30 degrees
104sin(30)=b'
51 km/h=b
160-51=109 -- I expected the answer to be 40 km/h, because i thought 4km was the horizonatl distance, so 160-20...

 

[BobG]

ok, here is what i thought, i used related rates:
triangle with sides a (height of plane=2), b (sqrt12, found by pythag.theorem) and side c (hypotenuse, =4 km, ditsance from car-plane)
c'=(sqrt12*120)/4
=104
then to find theta
sintheta=2/4, theta = 30 degrees
104sin(30)=b'
51 km/h=b
160-51=109 -- I expected the answer to be 40 km/h, because i thought 4km was the horizonatl distance, so 160-20...

Your c' is not close to being correct.

Symbolically, if y is the hypotenuse and x is the horizontal component, then:

y=\sqrt{2^2+x^2}

Differentiate to find dy. Then substitute in the values you know. You know dy=-160. You know x = \sqrt{12}
The only unknown is dx.

Keep in mind that dx has two things affecting it: the speed of the plane and the speed of the car.

 

[mathman100]

i'm not sure i understand- do i have enough information to sove if i know thta:
a=2, da/dt=0
b=sqrt12, db/dt=?
c=4, dc/dt= -160km/h
then where do i use the speed of plane=120km?

 

[BobG]

i'm not sure i understand- do i have enough information to sove if i know thta:
a=2, da/dt=0
b=sqrt12, db/dt=?
c=4, dc/dt= -160km/h

Have you differentiated your equation for the Pythagorean Theorem yet? Try doing that (you have to use the Chain Rule, which is what eventually results in a dx showing up in your equation).

then where do i use the speed of plane=120km?

Keep in mind that dx has two things affecting it: the speed of the plane and the speed of the car.

 

[mathman100]

Here's what i have:
triangle xyz where x=sqrt12 (by using pyth.theorem), y=2, and z=4.
dx/dt=? , dy/dt=0 , dz/dt=-160?
x^2+y^2=z^2
x(dx/dt)+y(dy/dt)=z(dz/dt)
since dy/dt=0:
x(dx/dt)=z(dz/dt) *** This is where i get stuck again- i want dx/dt, but where do i put in the speed of the plane, 120km/h?
If i just did it as is, then:
sqrt12(dx/dt)=4(-160)
dx/dt=-640/sqrt12
=about-184.8, which i can tell isn't right... What do i do??

 

[civil_dude]

I don't think the elevation difference matters, and that this is a simple relative velocity problem.

Vplane =Vcar + Vp/c

120 = Vc + 160

So Vc = -40, i.e. the car is going 40km/h east

I think..

 

[mathman100]

that's what i thought at first too, but i tried doing it the other way, and using that 185, you take away the speed of the plane (-120), so 185-120 =65km/hr--- the speed of the car i guess?

 

[mukundpa]

160 Km/h is the speed at which the distance between the car and the plane (y) is decreasing, not the horizontal distance x.

y^2 = x^2 + z^2

diff. we get 2y(dy/dt) = 2x(dx/dt) + 0

dy/dt is the rate at which y changes (160 Km/h) and dx/dt is the rate at which the horizontal distance x changes (120 + v).

M.P.

 

[mathman100]

thanks everyone for your help, you're awesome!