Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.6 m/s2 for 4.5 seconds. It then continues at a constant speed for 7.3 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 201 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

How fast is the blue car going 3.6 seconds after it starts? Answer: 12.96 m/s
How fast is the blue car going 8.8 seconds after it starts? Answer: 16.2 m/s
How far does the blue car travel before its brakes are applied to slow down? Answer: 154.71 m/s

These problems I had difficulty solving for. So any help is greatly appreciated. Thanks.

What is the acceleration of the blue car once the brakes are applied?
What is the total time the blue car is moving?
What is the acceleration of the yellow car?

at 3.6 seconds v = a t = 3.6*3.6 = 12.96 m/s

at 4.5 s
v = a t = 3.6*4.5 = 16.2 m/s
x(position) = d (distance traveled)
= average speed (16.2/2) * time (4.5)
= 8.1 m/s * 4.5 s = 36.45 meters

speed remains 16.2 until t = 4.5+7.3 = 11.8 seconds

at 4.5 + 7.3 = 11.8 seconds
v = 16.2 still
d = 16.2 * 7.3 = 118.26 m
so
x = 118.26 + 36.45 = 154.71 meters

at end
v = 0
so
average v = (16.2+0)/2 = 8.1 m/s
d = 201 - 154.71 = 46.29 meters to stop
time to stop = 46.29/8.1 = 5.71
acceleration while braking = (0-16.2)/5.71 = -2.84 m/s^2
so
total time = 5.71 + 11.8 = 17.5 s

so
yellow car goes 201 meters in 17.5 seconds
201 = a t^2 = a (17.5)^2
a yellow = .656 m/s^2

formula is wrong in the last problem. 1/2 is missing before a.

To find the acceleration of the blue car once the brakes are applied, we can use the formula:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (the velocity of the blue car just before the brakes are applied)
a = acceleration
s = distance travelled before braking (which we need to calculate)

We already have the value for v (0 m/s) and we know the initial velocity after 7.3 seconds is the same as the velocity just before braking, which is 16.2 m/s. We also know the distance travelled before braking is 201 meters. We can plug these values into the formula and solve for a:

0^2 = 16.2^2 + 2a(201)
0 = 262.44 + 402a
402a = -262.44
a = -262.44/402
a ≈ -0.652 m/s^2

Therefore, the acceleration of the blue car once the brakes are applied is approximately -0.652 m/s^2 (negative because it's deceleration or slowing down).

To find the total time the blue car is moving, we can add up the time it took for acceleration, the time it took for constant speed, and the time it took for deceleration:

time for acceleration = 4.5 seconds
time for constant speed = 7.3 seconds
time for deceleration = time for acceleration = 4.5 seconds

Total time = time for acceleration + time for constant speed + time for deceleration
Total time = 4.5 seconds + 7.3 seconds + 4.5 seconds
Total time ≈ 16.3 seconds

Therefore, the total time the blue car is moving is approximately 16.3 seconds.

To find the acceleration of the yellow car, we can use the formula:

v = u + at

Where:
v = final velocity (the velocity of the blue car at the end)
u = initial velocity (0 m/s, as the yellow car starts from rest)
a = acceleration of the yellow car (which we need to calculate)
t = time taken to catch the blue car (which is the same as the total time the blue car is moving)

We already know the final velocity of the yellow car is the same as the final velocity of the blue car, which is 0 m/s. We also know the initial velocity of the yellow car is 0 m/s, and the time taken to catch the blue car is approximately 16.3 seconds. We can plug these values into the formula and solve for a:

0 = 0 + a(16.3)
0 = 16.3a
a = 0/16.3
a = 0 m/s^2

Therefore, the acceleration of the yellow car is 0 m/s^2, meaning it moves at a constant velocity throughout the entire distance.

To find the answers to the remaining questions, we can break down the problem into smaller parts and apply the laws of motion.

1. What is the acceleration of the blue car once the brakes are applied?
To find the acceleration when the brakes are applied, we need to use the equation of motion:

v^2 = u^2 + 2as

Here, v is the final velocity (0 m/s since the car comes to rest), u is the initial velocity (the speed of the blue car at the moment brakes are applied), a is the acceleration (unknown), and s is the distance traveled before the brakes are applied (also unknown).

We know the initial velocity (u) of the blue car is the speed at 7.3 seconds, which we need to calculate:
u = a*t + u0
u = 3.6 m/s^2 * 7.3 s

Once we have the initial velocity (u), we can use the equation of motion to find the acceleration (a).

2. What is the total time the blue car is moving?
The total time the blue car is moving is the sum of the time it takes to accelerate and the time it moves at a constant speed. So, the total time is:

Total time = time to accelerate + time at constant speed
Total time = 4.5 s + 7.3 s

3. What is the acceleration of the yellow car?
Since the yellow car accelerates uniformly for the entire distance and catches up with the blue car right as it comes to a stop, we know that both cars travel the same distance in the same time. Therefore, the acceleration of the yellow car is also 3.6 m/s^2.