Rates of Change Question

A railroad bridge is 20m above, and at right angles to, a river. A person in a train travelling at 60 km/h passes over the centre of the bridge at the same instant that a person in a motorboat travelling at 20km/h passes under the centre of the bridge. How fast are the two people separating 10s later?

 

Please someone show me how to answer this!
Thanks!!

 

Only problem is I don't really know how to start!:(

 

ok this is what i tried...d^2= x^2 + y^2
took derivative
2d*dd/dt=2x*dx/dt + 2y*dy/dt
factored a 2 out and emilinated and isoltaed for dd/dt
dd/dt = (x*dx/dt + y*dy/dt)/d

 

subed in the values of the corresponding speeds 10s later
dd/dt=[(16.66)(166.6)+(5.55)(55.5)] /175.68
=17.5
and when i covert back to km/h i get 63.2 km/h
the answer is supose to be 63.2 km/h

 

opps the answer is supose to be 62.8 km/h

 

and I never rounded any if my numbers, stored them all in my calc.

 

yes I did convert.. the 60 km/h goes to 166.66m/s and the 20km/h to 5.55m/s

 

it's ok. I don't really see what I am doing wrong, I thought I was approaching the question right, but if my teacher says that is the answer then i guess not!

 

yeah maybe.. only problem is this is an assignment that's due for marks!!
Well thank you for your help!!:D

 

does the height of 20m have anything to do with the calculations?