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Expert: - 5/27/2008


Question
Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0.  After her chute opens, her velocity satisfies the differential equation dv/dt=-2(v+17), with initial condition v(0)=-47.  
Use separation of variables to finda an expression for v in terms of t, where t is measured in seconds.  
Terminal velocity is defined as limv(t) as t approaches infinity.  Find the terminal velocity of the skydiver to the nearest foot per second.  
It is safe to land when her speed is 20 feet per second.  At what time t does she reach this speed?

Answer
Note that when falling under gravity, the equation for velocity is
v = -9.8t since 9.8 is the gravitational constant.  You can set this equal to -47 and solve for when her chute opened.

The variable t is now reset to 0.  You must solve the equation
dv/(v+17) = -2dt using v(0)=47.  Set v equal to 20 and solve for t.  This value of t must be added to the value of t from the last equation to get the overall time.

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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