What is the terminal velocity and the time after which the velocity has decreased by 95% of the initial velocity in the following case:
The distance s (in m), fallen by a skydiver t seconds after jumping (and before the parachute opens) is s= 160(1/4-1+e^-t/4). Determine Vt = V(at t tends to infinity). This is the terminal velocity, the constant velocity attained when the air resistance balances the force of gravity. At what time has the velocity decreased by 95% of the initial velocity.
The distance s that the pilot falls after the parachute opens is given in terms of time as s = 160(1/4-1+e^-t/4)
Velocity is the derivative of s with respect to time and given as:
`(ds)/(dt) = 160*(-1/4)e^(-t/4) = -40*e^(-t/4) = -40/(e^(t/4))`
For t tending to infinity: `lim_(t->oo)(-40/e^(t/4)) = 0`
The terminal velocity of the skydiver which is the constant velocity attained when the air resistance balances the force of gravity is equal to 0 m/s.
The velocity when the pilot jumps off is equal to `lim_(t->0)(-40/e^(t/4)) = -40`
If the time at which the velocity is 5% of the initial velocity is t, `-40*e^(-t/4) = -(0.05*40)` .
Solving for t gives:
`e^(-t/4) = 0.05`
=> `-t/4 = ln 0.05`
=> t = 2.996 s.
At approximately 2.996 s after the pilot jumps off, the velocity is 95% of the initial velocity at jump off.