please somebdy help me solveing this.

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 4.3 m/s2 for 3.3 seconds. It then continues at a constant speed for 14.4 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 255.14 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.

1) How fast is the blue car going 7.3 seconds after it starts?
2) How far does the blue car travel before its brakes are applied to slow down?
3) What is the acceleration of the blue car once the brakes are applied?
4) What is the total time the blue car is moving?

To solve this problem, we will use the equations of motion for uniformly accelerated motion. We will break down the problem into different stages: acceleration, uniform motion, and deceleration. Let's solve each question step by step:

1) How fast is the blue car going 7.3 seconds after it starts?

To find the speed of the blue car at any given time, we need to use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In the first stage, the blue car accelerates uniformly at a rate of 4.3 m/s^2 for 3.3 seconds. Since it started from rest, the initial velocity, u, is 0.

v = 0 + (4.3 m/s^2)(3.3 s)
v = 14.19 m/s

Therefore, the blue car is going 14.19 m/s after 7.3 seconds.

2) How far does the blue car travel before its brakes are applied to slow down?

To find the distance traveled during uniform acceleration, we 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.

In this case, the blue car accelerates uniformly for 3.3 seconds. Similar to the previous question, the initial velocity, u, is 0.

s = (0)(3.3 s) + 0.5(4.3 m/s^2)(3.3 s)^2
s = 24.12 m

The blue car travels a distance of 24.12 meters before its brakes are applied.

3) What is the acceleration of the blue car once the brakes are applied?

The blue car applies the brakes and decelerates uniformly until it comes to rest. The equation for deceleration is the same as acceleration but with a negative sign:

v = u + at

In this case, the final velocity, v, is 0 (as the car comes to rest), the initial velocity, u, is the speed at the time the brakes are applied (which we need to find), and t is the time it takes for the car to come to rest (unknown).

We know the distance the car traveled before coming to rest, which is given as 255.14 meters.

Using the equation:

s = ut + (1/2)at^2

we can calculate the time it takes for the car to come to rest, t.

255.14 = u(14.4) + 0.5(-a)(14.4)^2

Simplifying this equation will give us the value of u.

After finding the value of u, we can use the equation v = u + at to find the acceleration a.

4) What is the total time the blue car is moving?

The total time the blue car is moving is the time it takes to accelerate uniformly (3.3 seconds), the time it travels at constant speed (14.4 seconds), and the time it decelerates until it comes to rest (which we will find in question 3).

The total time the blue car is moving is the sum of these three times.