Question

A weather balloon is released and rises vertically such that its distance s(t) above the ground during the first 10 seconds of the flight is given by s(t) = 7 + 2t + t^2, where s(t) is in feet and t is in seconds.

a) Find the velocity of the balloon at t=1, t=4
b) Find the velocity and acceleration (in proper units) at the instant the balloon is 51 feet above the ground.

Answer 1

are you familiar with derivatives?

Answer 2

Yes. Would the first thing I do is find s'(t)?

Answer 3

yep

Answer 4

s'(t)= 2t+2 What would I do after that?

Answer 5

find s'(1) and s'(4)

Answer 6

pretty easy...
for the second part..

Answer 7

Okay. For the second part I just do 51=7+2t+t^2 right? then equal to zero and find the roots?

Answer 8

solve for t when s(t) = 51
use that t (call it tf, for fun time) to find s'(tf) and s''(tf) (hint: s''(t), aka acceleration, is constant, so it doesn't depend on time anyway)

Answer 9

yes:) exactly.

Answer 10

Alright! Thanks a lot! :)

Answer 11

sure:) gl!

Answer 12

To find acceleration I just find the second derivative?

Answer 13

yes.