Question

At 1:59 PM a car leaves an intersection heading East at a constant rate of 40 ft/sec. One minute later, at 2 PM, a second car leaves the same intersection heading North at the constant rate of 100 ft/sec. Find the rate at which the distance between the two cars is changing at 2:01 PM.

Answer 1

128.1 ft/s

Explanation:

The first car at 1:59 PM moving at a rate of 40 ft/s heading East moves a distance of 40 ft/s × 1 s = 40 ft from 1:59 PM to 2:00 PM.

At the instant of 2:00 PM, the second car crosses the same intersection at a constant rate of 100 ft/s. From 2:00 PM to 2:01 PM which is 1 s, the first car moves a distance of 40 ft/s × 1s = 40 ft

The second car also moves a distance of 100 ft/s × 1 s = 100 ft.

So, the total distance moved by the first car at 2:01 PM is 40 ft + 40 ft = 80 ft and that moved by the second car is 100 ft.

Since their directions are perpendicular, we apply Pythagoras' theorem to find the distance between them at 2:01 PM.

So d = √(100² + 80²)

= √(10000 + 6400)

= √16400

= 128.06 ft

≅ 128.1 ft

So the rate at which the distance between the two cars is increasing at 2:01 PM is their distance at 2:01 PM divided by the time duration from 2:00 PM to 2:01 PM which their distance apart changed which is 1 s.

rate = d/t = 128.1 ft/1s = 128.1 ft/s