SOLUTION: A car travelling at 50 kilometer per hour crosses a bridge over a river 10 minutes before a boat travelling at 40 kilometers per hour passes under the bridge. The river and the bri

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Question 1157747: A car travelling at 50 kilometer per hour crosses a bridge over a river 10 minutes before a boat travelling at 40 kilometers per hour passes under the bridge. The river and the bridge are straight and at right angles to each other. At what rate are the car and the boat separating 10 minutes after the boat passes under the bridge if the height of bridge is 10 meters?
Answer by KMST(5404) About Me  (Show Source):
You can put this solution on YOUR website!
The speeds in km/minute are 50%2F60=5%2F6 for the car and 40%2F60=4%2F6=2%2F3 for the boat.
Defining the variable
t%22=%22time since the boat passed under the bridge, in minutes,
t%2B10%22=%22time since the car crossed the bridge, in minutes

Assuming that the road continues to be perpendicular to the river after the bridge,
for all t%3E=0, we have
x=+%285%2F6%29%28t%2B10%29}%22=%22distance between the car and the bridge, in km
y=+%284%2F6%29t%22=%22 =distance between the boat and the point under the bridge, in km
z=0.01%22=%22vertical distance between the boat and the bridge level
For all t%3E=0 ,
the coordinates for the car include y=0 and z=0 , assuming the road is level and continues to be perpendicular to the bridge, while
the boat coordinates include x=0, and z=1.
The differences in x, y, and z coordinates between car and boat are
DELTA%28x%29%22=%22%285%2F6%29%28t%2B10%29-0+=+%285%2F6%29%28t%2B10%29 ,
DELTA+%28y%29%22=%22%284%2F6%29t-0=%284%2F6%29t , and
DELTA+%28z%29%22=%220.01 ,
so the distance between the car and the boat, as a function of time is
%22=%22sqrt%28+%2825%2F36%29%28t%5E2%2B20t%2B100%29%2B%2816%2F36%29t%5E2%2B0.0001%29
%22=%22
%22=%22sqrt%28%2825%2F36%2B16%2F36%29t%5E2%2B%28500%2F36%29t%2B2500%2F36%2B0.0001%29%22=%22highlight%28sqrt%28%2841%2F36%29t%5E2%2B%28125%2F9%29t%2B625%2F9%2B0.0001%29%29%29
For t=10, that distance is
D%2810%29=sqrt%2841%2F36%29100%2B%28500%2F36%2910%2B2500%2F36%2B0.0001%29%22=%22sqrt%28%284100%2B5000%2B2500%29%2F36%2B0.0001%29%22=%22sqrt%2829000009%2F90000%29
%22=%22highlight%28sqrt%2829000009%29%2F300%29=approximately17.95055214

The rate of change of D%28t%29 as a function of t is the derivative for every value of t.

At t=10 , that is
%28%2841%2F18%29%2A10%2B125%2F9%29%2F%282D%2810%29%29%22=%22%2841%2A5%2F9%2B125%2F9%29%2F%282sqrt%2829000009%29%2F300%29%22=%22%28330%2F9%29%28300%2F2sqrt%2829000009%29%29%22=+approximately%22%28330%2A300%2F9%29%281%2F2sqrt%2829000009%29%29
%22=%2211000%2F%282sqrt%282900009%29%29%22km+%2F+minute%22%22=%22
approximately1.021324201492%22=%22%22km+%2F+minute%22
In Km/h, that is
%285500%2Fsqrt%282900009%29%29%2A60%22=%22highlight%28330000%2Fsqrt%2829000009%29%29%22km+%2F+h%22%29%22=%22approximatelyhighlight%2861%29+%0D%0A%22km+%2F+h%22%29