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

Answer:

t = 1.07 seg

Explanation:

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

$\frac{dv}{dt} = -2v-32$

This is a differential equation of separable variables

$\frac{dv}{-2v-32} = dt$

Multiplying by -1 to both sides of the equation

$\frac{dv}{2v+32} = -dt$

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

$\frac{ln(2v+32)}{2} = -t +k$

          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:

$-20 = -34e^{-2t}-16$

$\frac{-4}{-34} = e^{-2t}$

$ln(\frac{4}{34} ) = -2t$

t = 1.07 seg