2
$\begingroup$

A camera is located $100$ feet from a straight road along which a car is travelling at $120$ ft/sec. The camera turns so that it is always pointed at the car. In radians per seconds, how fast is the camera turning as the car passes closest to the camera?

I'm at a loss as to how to approach this problem... Does the car travel $100 \cdot x = 200 \ \text{feet}$ while the camera points at it? Then that would mean the total time interval would be $\frac{200}{12} = \frac{5}{3}$seconds, and the angle the camera turns would be $\pi$ total, but I don't see how I can find the speed at the specific point.

$\endgroup$
  • $\begingroup$ Write an equation showing the angle of the camera as a function of the distance of the car from the point closest to the camera. This will depend on the inverse tangent function. Next use that function to relate the rate of change of the car's distance from that closest point, to the rate of change of the angle of the camera. $\endgroup$ – samerivertwice Nov 28 '17 at 10:53
0
$\begingroup$

The following figure depicts the situation:

enter image description here

Assuming that the car hit the $0$ point at $t=0$ and is at $t^{th} $ second at the point $120t$ feet (the speed is $120$ feet/sec.) At the same moment the angle of the camera is

$$\alpha(t)=\operatorname{atan}\left(\frac65t\right).$$

The angular speed then equals $$\lim_{\Delta t\to 0}\frac{\operatorname{atan}\frac65(t+\Delta t)-\operatorname{atan}\left(\frac65t\right)}{\Delta t}=\frac t{dt}\operatorname{atan}\left(\frac65t\right)=\frac 65\frac1{1+\left(\frac56\right)^2t^2}.$$

The angular speed changes as shown below

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.