Related Rates Word Problem

O

ogg

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Nov 1, 2012
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Hello I have a question that I would love some help on. The question is:
Two cars leave an intersection, one 15 minutes after the other. The first car, A, travels north at the speed of 80 km/h and the second car, B, travels west at a speed of 60 km/h. How fast are the cars separating from each other 30 minutes after the first car leaves the intersection?

So far I have got:
d= (x^2+y^2)^.5
Because car 'A' leaves 15 minitue earlier I believe it would cause it to become ( y+20)^2
Total time than becomes 15 minutes of driving for both cars due to above compensation.
eqn becomes r^2= y^2 +40y+40+x^2
by taking the derivative I get 2r(dr/dt)= 2ydy/dt +40ydy/dt +2xdxdt
This i where I am stumped. I do not know what values to use for y,x,r. I know that we are solving for dr/dt and that dy/dt =80 and dx/dt=60

Any help would be great
Thanks!
 
Let the intersection be the origin of our coordinate system. Let distances be measured in kilometers and time in hours. I would let the rates of the two cars be expressed in kph, and using the given information, we may state the following two IVPs:

\(\displaystyle \dfrac{dx}{dt}=80\) where \(\displaystyle x(0)=0\)

\(\displaystyle \dfrac{dy}{dt}=-60\) where \(\displaystyle y\left(\frac{1}{4} \right)=0\)

Solving these, we find:

\(\displaystyle x(t)=80t\)

\(\displaystyle y(t)=-60t+15\)

Let \(\displaystyle D(t)\) represent the distance separating the cars, and by Pythagoras we may state:

\(\displaystyle D^2(t)=x^2(t)+y^2(t)\)

Differentiation with respect to \(\displaystyle t\) yields:

\(\displaystyle D'(t)=\dfrac{x(t)x'(t)+y(t)y'(t)}{D(t)}\)

Now, evaluate this at \(\displaystyle t=\dfrac{1}{2}\).
 
you have me confused here.
Woudlnt dy/dx be 80 as that is north?

and why is dy/dx at (1/4)= 0.

by doing it this way i got an answer of 95.97

By finishing it the way i started it i got 82.

anyone else have any input?

Thanks for your help I am just a little confused!
 
I should have used the positive y-axis as north and the negative x-axis as west, but it doesn't affect the outcome since we can use any variable names for our perpendicular axes. I agree with your decimal approximation of 95.97 kph.
 
Maybe think of it this way.

As car A is traveling at 80km/h and leaves 15 minutes (1/4 of an hour) earlier, therefore after 15 minutes has traveled 20km. Call this time zero. Another 15 minutes later has traveled a total of 40km.

Car B is traveling at 60km/h and travels for a total of 15 minutes, moving a total of 15km.

Therefore 30 minutes after car A leaves then x=15, y=40, r=42.72

This gives the equations x(t)=60t-15, y(t)=80t

Plugging in for dr/dt gives 95.97.

Don't know if that helps, but...
 
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