Math / Calculus

Question(1 point)
Suppose a police officer is $1 / 2$ mile south of an intersection, driving north towards the intersection at 35 mph . At the same time, another car is $1 / 2$ mile east of the intersection, driving east (away from the intersection) at an unknown speed. The officer's radar gun indicates 20 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 20 mph ). What is the speed of the other car?
Speed $=$ $\square$ mph.
Now suppose that the officer's radar gun indicates -20 mph instead (that is, the straight-line distance is decreasing at a rate of 20 $\mathrm{mph})$ . What is the speed of the other car this time?
Speed $=$ $\square$ mph.

Studdy Solution
For the second scenario, given $\frac{dd}{dt} = -20$ mph, substitute into the differentiated equation and solve for $v$ :
$-20 = \frac{1}{\sqrt{(\frac{1}{2} + vt)^2 + (-\frac{1}{2} + 35t)^2}} \left( -(\frac{1}{2} + vt)v + (-\frac{1}{2} + 35t)35 \right)$
Solve the equation for $v$ .
The speed of the other car in the first scenario is:
$\boxed{45 \text{ mph}}$
The speed of the other car in the second scenario is:
$\boxed{25 \text{ mph}}$

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