A spherical balloon is being inflated. Find the rate of increase of the surface area (S = 4Ï€r2) with respect to the radius r when r is each of the following. (Answers in unit ft2/ft) (a) 1 ft

(b) 3 ft

(c) 6 ft

A) 8π ft²/ft

B) 24π ft²/ft

C) 48π ft²/ft

Explanation:

Surface area of the spherical balloon is not clear here but it is supposed to be;

S = 4πr²

where:

r is the radius of the spherical balloon

So thus, the rate of change of the surface area of the spherical balloon by its radius will be:

dS/dr = 8πr

A) at r = 1ft;

dS/dr = 8 × π × 1

dS/dr = 8π ft²/ft

B) at r = 3 ft;

dS/dr = 8 × π × 3

dS/dr = 24π ft²/ft

C) at r = 6ft;

dS/dr = 8 × π × 6

dS/dr = 48π ft²/ft

anchor = archer //

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer