As soon as a traffic light turns green, a car speeds up from rest to 45.0 mi/h with constant acceleration 8.50 mi/h/s. In the adjoining bike lane, a cyclist speeds up from rest to 15.0 mi/h with constant acceleration 12.50 mi/h/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?

CAR:

V1 = 45mi/h * 1600m/mi * 1h/3600s=20 m/s
a=8.5mi/h/s*1600m/mi*1h/3600s=3.78m/s^2
a*t = 20 m/s
3.78t = 20
t = 5.29 s. to reach max velocity.
d1 = 0.5a*t^2
d1 = 0.5*3.78*5.29^2 = 52.9 m = Dist.@
which max. velocity is reached.

BIKE:
V2 = (15/45)*20 m/s = 6.67 m/s = max.
velocity.

a = (12.5/8.5)*3.78 = 5.56 m/s^2.

a*t = 6.67 m/s
5.56t = 6.67
t = 1.2 s to reach max velocity.

d2 = 0.5a*t^2
d2 = 0.5*5.56*1.2^2 = 4.0 m = Dist. @
which max. velocity is reached.

d1 = 0.5*3.78*1.2^2 = 2.72 m. = Dist.of
the car after 1.2 s.

D = d2-d1 = 4-2.72 = 1.28 m = Dist. the
by which the bike leads after 1.2 s.

When the car catches up:
a. d1 = d2 + 1.28 m.
0,5*a*t^2 = V2*t + 1.28
0.5*3.78*t^2 = 6.67*t+1.28
1.89t^2-6.67t-1.28 = 0
t = 3.71 s.(Use Quadratic Formula).

b. D = d2-d1 = 4-2.72 = 1.28 m.

a. t = (V-Vo)/a = (15-0)/12.5 = 1.2 s.

b. a=12.5mi/h/s*1600m/mi*1h/3600s=5.56m/s^2
= Acceleration of bike.
Db = 0.5*5.56*1.2^2 = 4 m. = Dist. of bike after 1.2 s.

a = (8.5/12.5) * 5.56 = 3.78 m/s^2 =
Acceleration of the car.
Dc = 0.5*3,78*1.2^2 = 2.72 m. = Dist. of the car after 1.2 s.

D = Db - Dc = 4 - 2.72 = 1.28 m

To answer these questions, we can use the equations of motion. Let's denote the time interval for which the bicycle is ahead of the car as "t". Now let's solve each part of the problem step by step.

(a) For what time interval is the bicycle ahead of the car?

We can calculate the time it takes for the car to reach a velocity of 45.0 mi/h using the equation:

vf = vi + at

Where:
vf = final velocity = 45.0 mi/h
vi = initial velocity = 0 (since the car starts from rest)
a = acceleration = 8.50 mi/h/s
t = time

Rearranging the equation to solve for t, we have:

t = (vf - vi) / a

t = (45.0 mi/h - 0) / 8.50 mi/h/s

t = 5.29 seconds (approx)

Now let's find the distance traveled by the bicycle during this time interval. We can use the equation:

d = vit + 0.5at^2

Where:
d = distance
vi = initial velocity = 0 (since the bicycle starts from rest)
a = acceleration = 12.50 mi/h/s
t = time (which we found to be 5.29 seconds)

Substituting the values into the equation, we get:

d = 0 + 0.5 * 12.50 mi/h/s * (5.29 s)^2

d ≈ 83.28 mi

Therefore, the bicycle is ahead of the car for approximately 83.28 miles during this time interval.

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

To find the maximum distance by which the bicycle leads the car, we need to calculate the distance traveled by the car during the same time interval.

Again using the equation:

d = vit + 0.5at^2

Where:
vi = initial velocity = 0 (since the car starts from rest)
a = acceleration = 8.50 mi/h/s
t = time (which we found to be 5.29 seconds)

Substituting the values into the equation, we get:

d = 0 + 0.5 * 8.50 mi/h/s * (5.29 s)^2

d ≈ 71.29 mi

Therefore, the maximum distance by which the bicycle leads the car is approximately 83.28 mi - 71.29 mi = 11.99 mi.

To answer this question, we need to determine the time interval for which the bicycle is ahead of the car and the maximum distance by which the bicycle leads the car.

First, let's calculate the time it takes for each vehicle to reach its cruising speed.

For the car:
Initial velocity, u = 0 mi/h
Final velocity, v = 45.0 mi/h
Acceleration, a = 8.50 mi/h/s

We can use the formula of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Rearranging the formula, we get:
t = (v - u) / a

Substituting the values, we have:
t_car = (45.0 - 0) / 8.50
t_car = 5.29 seconds (rounded to two decimal places)

For the bicycle:
Initial velocity, u = 0 mi/h
Final velocity, v = 15.0 mi/h
Acceleration, a = 12.50 mi/h/s

Using the same formula, we can calculate the time taken by the bicycle:
t_bike = (15.0 - 0) / 12.50
t_bike = 1.20 seconds (rounded to two decimal places)

(a) To find the time interval for which the bicycle is ahead of the car, we need to compare their respective times.

Since the bicycle takes a shorter time to reach its cruising speed, it is ahead of the car for that duration. Therefore, the bicycle is ahead of the car for t_bike = 1.20 seconds.

(b) To determine the maximum distance by which the bicycle leads the car, we need to calculate the distance traveled by each vehicle during the time t_car.

Using the formula of motion: s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

For the car:
Distance traveled, s_car = 0 * t_car + (1/2) * 8.50 * t_car^2
s_car = 0 + (1/2) * 8.50 * (5.29)^2
s_car = 74.80 mi (rounded to two decimal places)

For the bicycle:
Distance traveled, s_bike = 0 * t_car + (1/2) * 12.50 * t_car^2
s_bike = 0 + (1/2) * 12.50 * (1.20)^2
s_bike = 9.00 mi (rounded to two decimal places)

Therefore, the maximum distance by which the bicycle leads the car is s_bike - s_car = 9.00 - 74.80 = -65.80 mi (rounded to two decimal places).

Note: The negative sign indicates that the bicycle is behind the car at this point, as it takes less time to reach its cruising speed.