Car and bike physics problem?

As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration 9.00 mi/h 鈥?s. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration 13.0 mi/h 鈥?s. Each vehicle maintains constant velocity after reaching its cruising speed. (a) For what time interval is the bicycle ahead of the car? (b) By what maximum distance does the bicycle lead the car?

This is a straightforward application of the two formula for constant acceleration:

Vf = Vi + AT
D = Vi + (1/2)AT^2

where:
T is the time interval
A is the constant acceleration
Vi is the velocity at the beginning of the interval
Vf is the velocity at the end of the interval
D is the distance traveled during the interval.

The tricky part is keeping track of the intervals.

Since the bicyclist has the higher acceleration and lower top speed, the first interval is from the start (T = 0) to Tb, the time at which the bicyclist hits his top speed.

In this case, you have Vi, Vf, and A so you can compute Tb with the velocity equation.
You can correspondingly compute Cb and Bb, the distance traveled by the car and the bicycle during this interval, using the distance equation.

Next you can compute Tc, the time at which the car stops accelerating, using the velocity; Cc the distance traveled by the car to this point using the distance equation; and Bc, the distance traveled by the bicycle to this time as the sum of Bb and the cyclist's speed times (Tc - Tb).

If the car has passed the cyclist by this time, then you have to find the exact time in the interval at which this happened.

If that time is Tx, you would have the equation equating the two distances at time Tx:

Dc = (1/2)Ac(Tx)^2
Db = Bb + Vb(Tx - Tb)

This gives you a quadratic in Tx which you can solve for the smallest root greater than Tb.

If the car hasn't passed the cyclist by time Tc, then the cross-over time is in the interval when both are traveling at constant velocity so:

Db = Bb + Vb(Tx - Tb) = Bc + Vb(Tx - Tc)
Dc = Cc + Vc(Tx - Tc)

setting Db = Dc gives you a linear equation in Tx.

That takes care of part a.

For part b, the distance between the two keeps increasing up until the velocity of the car equals that of the cyclist. That happens between Tb and Tc.

So use the velocity equation with Vf = the final bicycle velocity, Vi = 0, and A = car's acceleration to get the time at which the gap is at its peak.

Then use the distance equations above to compute the corresponding distance.