Related Rates: Calculating the Separation Between Two Moving Objects

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A railroad bridge is 20m above a river, with a train traveling at 60 km/h and a motorboat at 20 km/h passing each other at the bridge. The problem involves calculating the rate at which the two are separating 10 seconds later using related rates. The separation distance formula d^2 = x^2 + y^2 is derived, leading to the equation dd/dt = (x*dx/dt + y*dy/dt)/d. After substituting values, the calculated separation rate is approximately 62.8 km/h, with the height of the bridge not affecting the calculations since it remains constant. The discussion emphasizes the importance of focusing on changing quantities in related rates problems.
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A railroad bridge is 20m above, and at right angles to, a river. A person in a train traveling at 60 km/h passes over the centre of the bridge at the same instant that a person in a motorboat traveling at 20km/h passes under the centre of the bridge. How fast are the two people separating 10s later?
 
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Please someone show me how to answer this!
Thanks!
 
Did you draw a picture? How does it look like?

Show your work please!
 
Only problem is I don't really know how to start!:(
 
Picture an aerial view. At some instance, the motorboat and train are at the same point. As they move further from each other (at right angles), what does their relationship start to resemble?

A triangle, rectangle, circle?
 
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
 
It asks you to evaluate it at 10s. You're rates have units km/h.

Convert hours to seconds.
 
  • #10
and I never rounded any if my numbers, stored them all in my calc.
 
  • #11
yes I did convert.. the 60 km/h goes to 166.66m/s and the 20km/h to 5.55m/s
 
  • #12
sabrina_m33 said:
yes I did convert.. the 60 km/h goes to 166.66m/s and the 20km/h to 5.55m/s
Sorry, didn't notice since you didn't write them in your problem.
 
  • #13
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!
 
  • #14
I got the same answer. Perhaps your teacher rounded off, who knows? Wouldn't worry too much.
 
Last edited:
  • #15
yeah maybe.. only problem is this is an assignment that's due for marks!
Well thank you for your help!:D
 
  • #16
Anytime, gl! Make 100s :)
 
  • #17
does the height of 20m have anything to do with the calculations?
 
  • #18
I got the same answer. and ur question no the 20m has nothing to do with the process for solving the question because the height between the bridge and the river does not change right ? so u do not need to differentiate that

Remember Related rates is only concerned with QUANTITIES THAT CHANGE ! lol sorry... i had to emphasize a point. This is what i learned from my hard times doing related rates and and... does anyone get another answer ? i got the same thing I am using Advanced FUnctions and Introductory Calculus from Harcourt Mathematics 12 . Thats where u got this question from lol and no from ur teacher llllooool <_<
 

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