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[Darya]

A man is standing 12 m east from the intersection point. A car is driving with velocity of 4 m/s north from that point. At one moment, the car is 9 m from the int.point. What's the rate of change of the distance between the man and the car at that point?

I didn't have a hard time solving this problem but there's a thing I can't still comprehend intuitively. So the rate of change, as we figure with calculus is 12/5 m/s. But if we change the distance of that car from the int. point, say 5 m, we get 12-5-13 sided triangle, velocity of the car left unchanged, then the rate of change of the distance between the man and the car is different now - 20/13. Why is it that way? The man isn't moving and the car is moving at the same speed, consequently, the distance should be changing at the same rate.

How can I view this problem better?

Thanks!!!

 

[Li Batton]

If the car starts 5 meters to the right of the intersection point and the man is still twelve meters east of it,
then that leg of the triangle becomes 12 + 5 = 17 meters. Try solving this and see what you get.

 

[Li Batton]

Also if you look at dr/dt, the rate the hypotenuse changes, is related to the distance y and the hypotenuse length.
so the hypotenuse grows at a different rate. dr/dt= (y/r)dy/dt. It seems counter intuitive but the square
of the hypotenuse grows as the squares of the other two legs. I hope this helps you.

 

[Darya]

Also if you look at dr/dt, the rate the hypotenuse changes, is related to the distance y and the hypotenuse length.
so the hypotenuse grows at a different rate. dr/dt= (y/r)dy/dt. It seems counter intuitive but the square
of the hypotenuse grows as the squares of the other two legs. I hope this helps you.

It kinda did.. Thanks!

 

[Li Batton]

Your welcome. It might also help to go back to the definition of
velocity to help you visualize what is happening. Velocity is
distance divided by time. So at that instant if your distances are
different then the velocity may not be the same.