- #1
mikiritenshi
- 5
- 0
I blanked out. Any help is greatly appreciated.
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: dv/dt= -2v-32, with initial condition v(0)=-50.
(a) Use separation of variables to find an expression for v in terms of t, where t is measured in seconds.
I got up to this far:
integral [dv/(-2v-32)]= integral (dt)
u= -2v-32; du= -2dv.
Help.
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: dv/dt= -2v-32, with initial condition v(0)=-50.
(a) Use separation of variables to find an expression for v in terms of t, where t is measured in seconds.
I got up to this far:
integral [dv/(-2v-32)]= integral (dt)
u= -2v-32; du= -2dv.
Help.