Car moving down road accelerates at uniform rate 2.8 m/s/s for 4.63 s to reach a speed of 106.7 m/s.

1) How fast was the car initially moving?
2) What distance did it travel during the a/c/c.?

This is straightforward stuff...

vf= vi + a*t solve for vi
d= vi*t + 1/2 a t^2

To answer these questions, we need to use equations of motion. There are three equations of motion that relate displacement (d), initial velocity (v₀), final velocity (v), acceleration (a), and time (t):

1) v = v₀ + at
2) d = v₀t + (1/2)at²
3) v² = v₀² + 2ad

Let's use the given information to find the answers:

1) How fast was the car initially moving?
In this case, we know the final velocity (v) is 106.7 m/s, acceleration (a) is 2.8 m/s², and the time (t) is 4.63 s. We can substitute these values into equation 1 and solve for v₀:

106.7 m/s = v₀ + (2.8 m/s²) * (4.63 s)

To isolate v₀, we bring (2.8 m/s²) * (4.63 s) to the other side of the equation as - (2.8 m/s²) * (4.63 s):

v₀ = 106.7 m/s - (2.8 m/s²) * (4.63 s)

By calculating this expression, we can find the initial velocity of the car.

2) What distance did it travel during the acceleration period?
To answer this question, we can use equation 2. We already know the initial velocity (v₀ = calculated from question 1), acceleration (a = 2.8 m/s²), and time (t = 4.63 s). We can substitute these values into equation 2 and solve for d:

d = v₀ * t + (1/2) * a * t²

By plugging in the calculated value for v₀, we can find the distance traveled during the acceleration period.

Using these equations and the given data, we can find the answers to the questions.