Hello, JaredMims!

A balloon is rising vertically above a level road at a constant rate of 1 ft/sec.

Just when the balloon is 65 feet above the ground,

a bicycle moving at a constant rate of 17 ft/sec passes under it.

(a) What is the distance between the balloon and the bicyclist three seconds later? Code:

A *
|\
t | \
| \
C * \
| \ d
| \
65 | \
| \
| \
D * - - - - * B
17t

The balloon is at $\displaystyle C$, rising at 1 ft/sec.

In the next $\displaystyle t$ seconds, it rises $\displaystyle t$ feet to $\displaystyle A.$

. . $\displaystyle AD \,=\,t + 65$ feet.

The cyclist is at $\displaystyle D$, moving at 17 ft/sec.

In the next $\displaystyle t$ seconds, it moves $\displaystyle 17t$ feet to $\displaystyle B.$

. . $\displaystyle DB \,=\,17t$

In 3 seconds, the balloon is 68 feet high, the cyclist moved 51 feet.

Pythagorus give us the distance: .$\displaystyle d \:=\:\sqrt{68^2+51^2} \:=\:\sqrt{7225} \:=\:85\text{ feet.}$

(b) What is the distance between the balloon and the bicyclist after another 3 seconds have passed?

After 6 seconds, the balloon is 71 feet high, the cyclist moved 102 feet.

Pythagorus says: .$\displaystyle d \:=\:\sqrt{71^2 + 102^2} \:\approx\:126\text{ feet}$