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

(a) First, we need to find the time it takes for each vehicle to reach its cruising speed.

For the car: The formula to find time when initial velocity (u), final velocity (v) and acceleration (a) are given is t = (v - u) / a. Here, u = 0 (since the car starts from rest), v = 50 mi/h and a = 9 mi/h/s. So, t_car = (50 - 0) / 9 = 5.56 hours.

For the bicycle: Using the same formula, with u = 0, v = 20 mi/h and a = 13 mi/h/s, we get t_bike = (20 - 0) / 13 = 1.54 hours.

The bicycle is ahead of the car for the time it takes the car to reach its cruising speed, which is 5.56 hours.

(b) To find the maximum distance the bicycle leads the car, we need to find the distance each vehicle travels until the car reaches its cruising speed.

The formula to find distance when initial velocity (u), time (t) and acceleration (a) are given is d = ut + 0.5at^2.

For the car: d_car = 05.56 + 0.59*(5.56)^2 = 139.05 miles.

For the bicycle: Since the bicycle reaches its cruising speed before the car, we need to calculate the distance in two parts - while it's accelerating, and while it's at constant speed.

While accelerating: d_bike1 = 01.54 + 0.513*(1.54)^2 = 15.38 miles.

While at constant speed: d_bike2 = 20 * (5.56 - 1.54) = 80.4 miles.

So, the total distance the bicycle travels is d_bike = d_bike1 + d_bike2 = 15.38 + 80.4 = 95.78 miles.

Therefore, the maximum distance the bicycle leads the car by is d_bike - d_car = 95.78 - 139.05 = -43.27 miles.

However, distance cannot be negative. This means that the bicycle never leads the car. The car overtakes the bicycle before it reaches its cruising speed.