Hi guys, would appreciate some help with this problem:

When a light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 8.80 mi/h·s. In the next bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 14.5 mi/h·s. Each vehicle maintains a constant velocity after attaining 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

Write equations for position vs time for both car and bike.

Solve for the time that the car passes the bike.
In other words, solve
X(car) = X(bike)
That will answer part (a).

For part (b), differentiate
X(bike) - X(car) vs. time and find out where the derivative is zero. That will be the time of maximum separation.

Show your work if you need additional help

First make everything feet and seconds

1 mi = 5280 ft
1 h = 3600 s
car
50 mi/h *5280 ft/mi *1h/3600s
= 73.3 ft/s
8.8 mi/hs *5280 ft/mi *1h/3600s =12.9ft/s^2
bike
20 mi/hr --> 29.3 ft/s
14.5 mi/hs --> 21.3 ft/s^2
Now as long as the bike is accelerating the bike will be getting further ahead. However once the car finally reaches 20 mph, the car will start to catch up.
So question (b) is easy. When does the car reach 20 mph or 29.3 ft/s ?
29.3 = 12.9 t
t = 2.27 s for the car to reach max bike speed, from then on it will be catching up.
How far did the car go before it started to catch up?
d = (1/2) (12.9) (2.27)^2=
33.2 ft (answer part b)
now when does the car finally pass the bike?

How long does the car accelerate?
73.3 = 12.9 t
t =5.68 s
during acceleration the car goes
d = .5 (12.9)(5.68^2) = 208 ft
from then on the car goes at 73.3 ft/s

How long does the bike accelerate?
29.3 = 21.3 t
t = 1.38 s
d = .5 (21.3)(1.38^2) = 20.1 ft
from then on the bike goes at 29.3 ft/s

Now probably the car passes the bike while the car is still accelerating so try that possibility first
d car = .5*12.9 * t^2
d bike = 20.1 + 29.3 (t-1.38)
6.45 t^2 = -20.3 +29.3 t
6.45 t^2 -29.3 t +20.3 = 0
t = 3.69 or .863
at .863 s the bike is ahead of the car and gaining, so that does not work
However at 3.69 s the car has gone
d = (1/2) (12.9) 3.69^2 = 87.8 ft
and the bike has gone
d = 20.1 + 29.3 (3.69-1.38) = 87.8 ft
so the bike was ahead for 3.69 s (part a)

To solve this problem, we can use the equations of motion for uniformly accelerated motion.

Let's start by finding the time it takes for each vehicle to reach their cruising speeds.

For the car:
Initial velocity of the car (u) = 0 mi/h
Final velocity of the car (v) = 50.0 mi/h
Acceleration of the car (a) = 8.80 mi/h·s

Using the equation v = u + at, we can solve for the time (t) it takes for the car to reach its cruising speed:

v = u + at
50.0 = 0 + 8.80t
8.80t = 50.0
t = 50.0 / 8.80
t ≈ 5.68 s

So it takes the car approximately 5.68 seconds to reach its cruising speed.

Similarly, for the bicycle:
Initial velocity of the bicycle (u) = 0 mi/h
Final velocity of the bicycle (v) = 20.0 mi/h
Acceleration of the bicycle (a) = 14.5 mi/h·s

Using the same equation v = u + at, we can solve for the time (t) it takes for the bicycle to reach its cruising speed:

v = u + at
20.0 = 0 + 14.5t
14.5t = 20.0
t = 20.0 / 14.5
t ≈ 1.38 s

So it takes the bicycle approximately 1.38 seconds to reach its cruising speed.

(a) To find the time for which the bicycle is ahead of the car, we need to compare the times it takes for each vehicle to reach their cruising speeds. In this case, the bicycle takes less time to reach its cruising speed, so the time for which the bicycle is ahead of the car is equal to the time it takes for the bicycle to reach its cruising speed, which is approximately 1.38 seconds.

(b) To find the maximum distance by which the bicycle leads the car, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

For the car:
s_car = (0)(5.68) + (1/2)(8.80)(5.68^2)
s_car = 0 + (1/2)(8.80)(32.2624)
s_car ≈ 142.28 ft

For the bicycle:
s_bicycle = (0)(1.38) + (1/2)(14.5)(1.38^2)
s_bicycle = 0 + (1/2)(14.5)(1.9044)
s_bicycle ≈ 19.51 ft

Therefore, the maximum distance by which the bicycle leads the car is approximately 19.51 ft.

To solve this problem, we will first find the time it takes for each vehicle to reach their cruising speed and then use that information to determine the time the bicycle is ahead of the car and the maximum distance by which the bicycle leads the car.

Given:
Car's initial velocity (u1) = 0 mi/h
Car's final velocity (v1) = 50.0 mi/h
Car's acceleration (a1) = 8.80 mi/h·s

Bicycle's initial velocity (u2) = 0 mi/h
Bicycle's final velocity (v2) = 20.0 mi/h
Bicycle's acceleration (a2) = 14.5 mi/h·s

Let's start with part a) to find the time the bicycle is ahead of the car:

The time it takes for the car to reach its cruising speed can be found using the formula:

v1 = u1 + a1 * t1

where v1 is the final velocity, u1 is the initial velocity, a1 is the acceleration, and t1 is the time taken.

Substituting the given values:
50.0 = 0 + 8.80 * t1

Simplifying:
50.0 = 8.80 * t1

Dividing both sides by 8.80:
t1 = 50.0 / 8.80
t1 = 5.68 seconds

Similarly, we can find the time it takes for the bicycle to reach its cruising speed:

v2 = u2 + a2 * t2

where v2 is the final velocity, u2 is the initial velocity, a2 is the acceleration, and t2 is the time taken.

Substituting the given values:
20.0 = 0 + 14.5 * t2

Simplifying:
20.0 = 14.5 * t2

Dividing both sides by 14.5:
t2 = 20.0 / 14.5
t2 = 1.38 seconds

Now that we have the time for each vehicle to reach its cruising speed, we can determine if the bicycle is ahead of the car and for how long.

Since the bicycle takes less time (1.38 seconds) to reach its cruising speed compared to the car (5.68 seconds), the bicycle is initially ahead of the car.

To find the time the bicycle is ahead of the car, we subtract the time it takes for the bicycle to reach its cruising speed from the time it takes for the car to reach its cruising speed:

Time Bicycle is ahead of Car = t1 - t2
Time Bicycle is ahead of Car = 5.68 seconds - 1.38 seconds
Time Bicycle is ahead of Car = 4.30 seconds

So, for part a) the bicycle is ahead of the car for 4.30 seconds.

Moving on to part b) to find the maximum distance by which the bicycle leads the car:

We can use the equation of motion:

s = u * t + 0.5 * a * t^2

where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.

For the car:
s1 = 0 * t1 + 0.5 * a1 * t1^2
s1 = 0 + 0.5 * 8.80 * (5.68)^2
s1 = 113.31 mi

For the bicycle:
s2 = 0 * t2 + 0.5 * a2 * t2^2
s2 = 0 + 0.5 * 14.5 * (1.38)^2
s2 = 13.71 mi

To find the maximum distance by which the bicycle leads the car, we subtract the distance traveled by the car from the distance traveled by the bicycle:

Maximum Distance Bicycle leads Car = s2 - s1
Maximum Distance Bicycle leads Car = 13.71 mi - 113.31 mi
Maximum Distance Bicycle leads Car = -99.60 mi

As we can see, the calculated result is negative, which indicates that the bicycle does not lead the car by distance. In fact, the car leads the bicycle by a distance of 99.60 mi.

Therefore, for part b) the car leads the bicycle by 99.60 mi.