Posted by **blair** on Friday, October 9, 2009 at 12:24am.

Vroom-vroom! As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 8.40 mi/h·s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 11.5 mi/h·s. Each vehicle maintains a constant velocity after reaching its cruising speed.

(a) For how long is the bicycle ahead of the car?

____________ s

(b) By what maximum distance does the bicycle lead the car?

______________ ft

Please explain how to solve this with the the necessary equations. Thanks!

- physics -
**MathMate**, Friday, October 9, 2009 at 12:10pm
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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