A car is traveling along a straight road at a velocity of +38.8 m/s when its engine cuts out. For the next 2.01 seconds, the car slows down, and its average acceleration is . For the next 6.36 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 8.37-second period is +26.6 m/s. The ratio of the average acceleration values is = 1.45. Find the velocity of the car at the end of the initial 2.01-second interval.

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To find the velocity of the car at the end of the initial 2.01-second interval, we can use the kinematic equation:

vf = vi + at

Where:
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration
- t is the time interval

Let's break down the given information:

Initial velocity (vi): +38.8 m/s
Time interval (t1): 2.01 seconds
Initial acceleration (a1): Unknown

We are given that the car slows down during this first interval, so the acceleration will be negative.

The velocity at the end of this interval (vf1) is not given and needs to be determined.

Now, using the given information about the second interval:

Time interval (t2): 6.36 seconds
Final velocity of the car at the end of the 8.37-second period (vf): +26.6 m/s
Acceleration ratio (a2 / a1): 1.45

Let's call the acceleration during the second interval a2.

Now, let's find the acceleration during the first interval (a1):

a2 / a1 = 1.45

We can rewrite this equation as:

a2 = 1.45 * a1

Now, let's substitute the values into the kinematic equation for the second interval:

vf = vi + a2 * t2

26.6 = vf1 + (1.45 * a1) * 6.36 ---(Equation 1)

And for the first interval:

vf1 = vi + a1 * t1

In this equation, we have three variables: vf1, a1, and vf. We need to solve Equation 1 for the variable vf1 and then substitute it into the above equation.

Rearranging Equation 1, we have:

vf1 = 26.6 - (1.45 * a1 * 6.36) ---(Equation 2)

Now, substitute Equation 2 into the equation for the first interval:

vf1 = vi + a1 * t1

vf1 = 38.8 + a1 * 2.01

Substitute the value of vf1 from Equation 2 into this equation:

26.6 - (1.45 * a1 * 6.36) = 38.8 + a1 * 2.01

Now, solve this equation for a1:

26.6 - 91.62 * a1 = 38.8 + 2.01 * a1

Combine like terms:

-14.2 = 93.63 * a1

Divide both sides by 93.63:

a1 = -14.2 / 93.63

a1 ≈ -0.1519 m/s²

Now, substitute this value of a1 into Equation 2 to find vf1:

vf1 = 26.6 - (1.45 * (-0.1519) * 6.36)

vf1 ≈ 36.81 m/s

Therefore, the velocity of the car at the end of the initial 2.01-second interval is approximately +36.81 m/s.