Answer:
To find the velocity of the car at the end of the initial 1.67-second interval, you can use the equations of motion for uniformly accelerated motion. Let's denote the initial velocity as "vi," the time as "t," the acceleration as "a," and the final velocity as "vf."
The equation relating these variables is:
vf = vi + at
In the first 1.67-second interval:
Initial velocity (vi) = 38.8 m/s
Time (t) = 1.67 seconds
Acceleration (a1) = ?
We need to find the acceleration (a1) in this interval. We are given that the ratio of the average acceleration values is a1/a2 = 1.76. Therefore, we can express the second acceleration (a2) in terms of a1 as a2 = a1/1.76.
Now, we know the final velocity (vf) at the end of the 5.33-second period is +28.5 m/s, and we can calculate the second interval's acceleration using the same formula:
vf = vi + at
28.5 m/s = 38.8 m/s + a2 * 5.33 s
Now, let's solve for a2:
a2 = (28.5 m/s - 38.8 m/s) / 5.33 s
a2 = (-10.3 m/s) / 5.33 s
a2 ≈ -1.93 m/s²
Now that we have found a2, we can find a1 using the given ratio:
a1 / a2 = 1.76
a1 / (-1.93 m/s²) = 1.76
Now, solve for a1:
a1 ≈ 1.76 * (-1.93 m/s²)
a1 ≈ -3.40 m/s²
Now that we have the acceleration in the first interval, we can use it to find the velocity at the end of the initial 1.67-second interval:
vf = vi + at
vf = 38.8 m/s + (-3.40 m/s² * 1.67 s)
vf ≈ 38.8 m/s - 5.68 m/s
vf ≈ 33.12 m/s
So, the velocity of the car at the end of the initial 1.67-second interval is approximately +33.12 m/s.