Question

A balloon is at a height of 50 meters, and it is rising at the constant rate of 5m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10m/sec. How fast is the distance between the bicyclist and the balloon increasing 10 seconds later?

Answer 1

The distance between the bicyclist and the balloon is not increasing 10 seconds later.

Step-by-step explanation:

We can find the rate at which the distance between the bicyclist and the balloon is increasing by considering the positions and speeds of both objects.

Since the balloon is rising at a constant rate of 5 m/sec, after 10 seconds it will be at a height of 50 meters + (5 m/sec × 10 sec) = 100 meters.

The bicyclist is traveling at a constant speed of 10 m/sec, so after 10 seconds, they will have traveled a distance of 10 m/sec × 10 sec = 100 meters.

Therefore, the distance between the bicyclist and the balloon is increasing at a rate of 100 meters - 100 meters = 0 meters/sec.

Answer 2

After 10 seconds the balloon will be at the height of 100 meters ( 50 + 5 * 10 = 50 + 50 = 100 m ).
After 10 seconds the bicyclist will travel the distance of: 10 * 10 = 100 m
The distance between the bicyclist and the balloon was at the start 50 m
( at t = 0 s) and after 10 seconds ( at t = 10 s ) it will be:
d² = 100² + 100² = 10,000 + 10,000 = 20,000
d = √20,000 = 100√2 = 141.42 m
The rate:
v = (141.42 - 50)m / 10 s = 9.142 m/s
The distance is increasing at the rate of 9.142 m/s.