How Do You Derive the Velocity Equation for the Pitt Fall Ride at Kennywood?

[member 217843]

Homework Statement

The Pitt Fall is thrill ride at Kennywood that lifts passengers to a certain height, pauses for a few moments, and then drops the riders, causing them to free fall towards the ground before gradually applying breaks 79 ft above the ground. Assume there is a drag force, F=-bv, and the terminal velocity reached is 65 mph (29.1m/s). There are 16 passengers, each weighing 178 lbs, and the ride weighs 10,000 lbs. Determine the value of b. Derive the velocity equation as a function of time.

Homework Equations

I understand that you must use calculus, but I do not know how to derive the equation.

The Attempt at a Solution

I found that b=1962.6, and I started to derive the equation, but I am not certain where to go from here:
F=-bv
ma=-bv
m(dv/dt)=-bv
dv/v=-b(dt)/m
$$\int dv/v$$=-b/m$$\int dt$$

 

[housemartin]

First you got value of b wrong i guess. And net force on the ride its not just -bv, there is gravity too, otherwise this whole ride would have no meaning ;] Net force is F = mg - bv (if you choose your positive axis downward)
Terminal velocity means that F=ma=0 (a = 0) so velocity does not change: mg - bv =0 -> b = mg/v (use right units too - if g is in m/s^2, then v in m/s)
As for finding v in terms of time, just write:
F = ma = m(dv/dt) = mg - bv