[tntitansgirl]

4. Two points, A and B, are 275 ft apart. At a given instant, a balloon is released at B and rises vertically at a constant rate of 2.5 ft/sec, and, at the same instant, a cat starts running from A to B at a constant rate of 5 ft/sec.
a. After 40 seconds, is the distance between the cat and the balloon decreasing or increasing? At what rate?
b. Describe what is happening to the distance between the cat and the balloon at t= 50 seconds.

[colbtech]

if you draw this on graph paper, it might make sense.

start with A and B, 275ft apart, so the cat and the balloon are also 275ft apart.

after 40 seconds, the cat is now at 200ft (ie 75ft away from where the balloon started), but the balloon has risen ? (40s x 2.5ft).

Now using Pythagoras you can calculate the distance the balloon is away fro m the cat.

to solve b) the cat moves on a further 5ft towards B and the balloon rises a further ?

Hope this helps