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!

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.

To solve this problem, we can use the equations of motion:

1. v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. s = ut + 0.5at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Let's first find the time it takes for each vehicle to reach its cruising speed.

For the car:
Initial velocity (u) = 0 mi/h
Final velocity (v) = 50.0 mi/h
Acceleration (a) = 8.40 mi/h*s

Using the equation v = u + at, we can rearrange it to t = (v - u) / a and plug in the values:

t_car = (50.0 mi/h - 0 mi/h) / 8.40 mi/h*s
t_car = 5.95 s

Similarly, for the bicycle:
Initial velocity (u) = 0 mi/h
Final velocity (v) = 20.0 mi/h
Acceleration (a) = 11.5 mi/h*s

t_bike = (20.0 mi/h - 0 mi/h) / 11.5 mi/h*s
t_bike = 1.74 s

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

Since the bicycle reaches its cruising speed faster, it will be ahead of the car for the time it takes for the car to reach its cruising speed. Therefore, the bicycle is ahead for t_car seconds.

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

To find the distance traveled by each vehicle during the time the car takes to reach its cruising speed, we can use the equation s = ut + 0.5at^2.

For the car:
u_car = 0 mi/h
t_car = 5.95 s
a_car = 8.40 mi/h*s

s_car = (u_car * t_car) + (0.5 * a_car * t_car^2)
s_car = 0 + 0.5 * 8.40 mi/h*s * (5.95 s)^2
s_car = 0.5 * 8.40 mi/h*s * 35.4025 s^2
s_car = 148.83 mi

For the bicycle:
u_bike = 0 mi/h
t_bike = 5.95 s
a_bike = 11.5 mi/h*s

s_bike = (u_bike * t_bike) + (0.5 * a_bike * t_bike^2)
s_bike = 0 + 0.5 * 11.5 mi/h*s * (1.74 s)^2
s_bike = 0.5 * 11.5 mi/h*s * 3.0276 s^2
s_bike = 9.80 mi

The maximum distance the bicycle leads the car is equal to the difference in the distances traveled by each vehicle:
max_distance = s_bike - s_car
max_distance = 9.80 mi - 148.83 mi
max_distance = -139.03 mi (notice that the distance is negative, indicating that the car is actually ahead of the bicycle)

To convert this distance to feet, we can use the fact that 1 mi = 5280 ft:

max_distance = -139.03 mi * 5280 ft/mi
max_distance = -733,342.4 ft

Therefore, the maximum distance the bicycle leads the car is approximately 733,342.4 feet.