Isaac C. answered 07/07/15
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In the initial position, the police car and the red car are at two vertices of a right triangle with leg of 30 and 210 feet. The line of site between them is the hypotenuse S. We can express the radar information as meaning dS/dt = -100;
Let r be the speed of the red car which is what we want to know.
We can see that the length of the red car leg of the triangle is decreasing and is given by 210-rt. The police car leg remains constant at 30 feet.
The hypotenuse of the triangle at time t=0 can be found from S2 = 302 + 2102 or about 212 ft.
The distance between red car and the police car has the following relationship S2 = 302 + (210 - rt)2 Taking the derivative of both sides with respect to t, we find that 2S*ds/dt = 0 -2r(210 -rt). This expression is true for all values of t, so let's use the values we have for t = 0;
2*212*(-100) = -2(r)210 = 101 ft/second. Note that this is larger than the speed measured by the speed gun. This should be what we expect.