Question

A weather balloon is released and rises vertically such that its distance s (t) above the ground during the first 10 sec of flight is given by s(t)-6+2t+t2, where s(t) is in feet and t is in (a) Find the velocity at t- 4 seconds. (Worth 5 points) sec/Pt Answer (b) Find the instant the balloon is 50 feet above the ground.

Answer 1

Step-by-step explanation:

A weather balloon is released and rises vertically such that its distance s (t) above the ground during the first 10 sec of flight is given by :

$s(t)=6+2t+t^2$ ...(1)

(a) We need to find velocity at t = 4 seconds

Differentiate equation (1) wrt t.

$v=\dfrac{ds}{dt}\\\\v=\dfrac{d(6+2t+t^2)}{dt}\\\\v=2+2t$

At t = 4 seconds,

$v=2+2(4)\\\\=10\ ft/s$

At t = 4 seconds, the velocity is 10 ft/s.

(b) When the balloon is 50 feet above the ground,

$6+2t+t^2=50$

We need to find t.

$6+2t+t^2=50\\\\t^2+2t-50+6=0\\\\t^2+2t-44=0$

It is a quadratic equation,

t = 5.708 and t = -7.708

Neglecting negative time, the instant time is 5.708 seconds.