Posted by blair on Friday, October 9, 2009 at 12:24am.
In general, if
v(t)=velocity at time t
v0 = initial velocity, and
a = acceleration,
then
v(t)= v0 + at
S(t)=v0*t + (1/2)at²
Consider individually the displacement of the bicycle, S1, and the car, S2, as a function of time, t.
Each of the functions S1(t) and S2(t) can be separated into two parts, the acceleration phase, and the cruising phase.
It is probably more convenient to convert the accelerations to ft/s² than to work with mi/hr-s.
The acceleration phase is from t=0 to t=topspeed/acceleration, after which they both cruise.
So by creating the functions S1(t) and S2(t), you can equate the two to solve for the time where the car catches up with the cyclist.
By differentiating the function and equating the derivative to zero, you can find the maximum lead of the cyclist.
S(t) = S1(t)-S2(t)
S'(t) = S1'(t) - S2'(t) = 0
Solve for t=to, evaluate S(to).
If you need help, post any time.
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