a ballon is rising vertically above a level, straight road at a constant of 1 ft/sec . just when the b?

Question

a ballon is rising vertically above a level, straight road at a constant of 1 ft/sec . just when the balloom is 65 ft above the graund a bycicle moving at a constant rate of 17 fit /sec passes under it. how fast is the distance between the bycicle and balloom encreasing 3 sec later

Answer 1

Ok, what I understand, is that we need to calculate the distance 3 seconds later.

At first, when tha ballon is 65 ft above the ground, the bycicle is just down the ballon, so the distance between the ballon and the bycicle is 65 ft, but 3 seconds later :

Ballon : 1 ft/sec * 3 sec = 3 ft

Now the ballon is 65 + 3 ft above the ground

the bycicle : 17 ft/sec * 3 sec = 51 ft

So, let's make a graphic, we will find a right triangle :

D = sqrt((51)^2 + (68)^2) = 85 ft

Answer : 85 ft

Hope that helped you