Question

A car travels down a highway at 25 m/s. An observer stands 150 m from the highway.

(a) How fast is the distance from the observer to the car increasing when the car passes in front of the observer? Explain your answer without making any calculations.

(b) How fast is the distance increasing 20 s later?

Answer 1

Im not sure if im am wholly correct. But here goes...

(a) How fast is the distance from the observer to the car increasing when the car passes in front of the observer?
My answer is 0. If you plot a graph showing the distance from the car to the observer against time. You will notice that in the beginning, the car is far away from the observer, and the distance is getting smaller and smaller. The rate of change is a decreasing (in magnitude) negative number.
When the car comes to right in front of the observer, the distance from the car to the observer is 150m fixed and at that instant, the rate of change is 0.
Previously we noted that the rate of change is a decreasing in magnitude, naturally it will hit 0 at some point and then continue increasing in the positive direction.

(b)How fast is the distance increasing 20s later?

Construct a simple equation, where D is the distance and t is time.

\[D = \sqrt{150^{2} + (25t)^{2}}\]
\[D= \sqrt{22500 + 1375t ^{2}}\]
\[\frac{dD}{dt} =\frac{2750t}{ \sqrt{22500+1375t^{2}}}\]
Substitue t = 20
dD/dt= 72.7m/s^2