Question

Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t ≥ 0 . After her parachute opens, her velocity satisfies the differential equation dvdt=−2v−32 , with initial condition v(0) = −50 . It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?

Answer 1

t = 1.07 seg

Step-by-step explanation:

First we are going to solve the differential equation for the velocity:

`(dv)/(dt) = -2v-32`

This is a differential equation of separable variables

`(dv)/(-2v-32) = dt`

Multiplying by -1 to both sides of the equation

`(dv)/(2v+32) = -dt`

We integrate the left side with respect to the velocity and the right side with respect to time

`(ln(2v+32))/(2) = -t +k`

where k is a integration constant

ln(2v+32) = -2t + k

`2v +32 = e^(-2t+k)`

`2v + 32 = ce^(-2t)`

`v = ce^(-2t) -16`

We determine the constant c with the initial condition v(0) = -50

`-50 = ce^(-2(0)) -16`

-50 + 16 = c

c = -34

Then

`v(t) = -34e^(-27)-16`

When the velocity is -20 ft/s the time is:

`-20 = -34e^(-2t)-16`

`(-4)/(-34) = e^(-2t)`

`ln((4)/(34) ) = -2t`

t = 1.07 seg