Two cars are travelling towards a stop sign on two roads that make a 90 degree angle. The first car is 165 feet from the sign and is travelling at a rate of 25 feet per second. The second car is 110 feet from the sign and is travelling at a rate of 35 feet per second. Find the rate of change at the distance between the cars after one second.
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1 Answer

After time t seconds, the first car is 165-25t feet away from the sign and the other car is 110-35t feet from the sign. The two cars and the sign form a right triangle with legs equal to the aforementioned distances.

The hypotenuse of the triangle has length x=√((165-25t)²+(110-35t)²).


dx/dt=(-15950+3700t)/√(39325-15950t+1850t²) is the rate of change of the straight line distance between them.

When t=1, dx/dt=-12250/√25225=-12250/158.8238=-77.13 ft/s approx. The negative sign indicates a decreasing rate of change (speed).


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