the speen in feet per second at which a skidiver is falling

the speen in feet per second at which a skidiver is falling t seconds after jumping out of a plane, is approximated by the function S(t)=80(1-e^-0.2t)

1- How fast is the skydiver falling after 2 seconds?

2- How long does it take for the falling speed to reach 40 ft/sec?

3- there is a speed above which the skydiver cant fall. What is this speed (terminal speed)?

with the steps please

Solution

S(t) = 80(1-e^-0.2t)

1. S(2) = 80(1-e^-(.2*2)) =80(1-e^-.4) = 26.37 feet per second

2. 40=80(1-e^-.2t)

.5=1-e^-.2t

e^-.2t=.5

Takin ln on both sides

-.2t= ln.5

t=ln.5/-.2 = 3.5 sec