related rates problem?

Question

a race official is watching a racecar approach a finish line at the rate of 200 km/hr. Supposed the officla is sitting at the finish line, 20 m from the point where the car will cross, and let ø be the angle between the finish line and the official's line of sight to the car. At what rate is ø changing when the car crosses the finish line? give answers in rad/s.

Answer 1

See this figure: http://i7.tinypic.com/4hib50n.gif

(I find it almost impossible to solve problems like these without drawing a diagram of the situation.)

The distance, x, from the car to the finish line is related to the angle ø by

tan(ø) = x/20 so

x = 20 tan(ø)

dx/dt = 20 sec²(ø) dø/dt so

dø/dt = (dx/dt)/(20 sec²(ø))

You know dx/dt (it's -200 km/hr; negative because x is decreasing with time), and you want dø/dt when ø=0.

It would be useful to convert the car's speed from km/hr to m/s before using it in the above.

Answer 2

draw a diagram...

A
|
|
| y, dy/dt = 200km/hr
|
|
|___________C
B 20 m

in case u don't understand, ø should be the angle ACB, point B is the point where the car will cross
and y = distance between the car and the point
now u have this function:

tan (ø) = y / 20

dø/dt [sec(ø)]^2 = 1/20 * dy/dt

now, ø=0 when the car just crosses the finish line, and dy/dt is given, all you need is to put in the value with the correct units

dø/dt * [sec(0)]^2 = 1/20m * 200000m/3600s
dø/dt = 2.778 rad/s