Question

A car starts from rest and accelerates at a rate of 5.75 m/s/s for a time (t1) of 6.76 seconds. The car then maintains a constant speed for a time of 7.32 seconds (t2 minus t1). Finally, the car slows down at a rate of -4.21 m/s/s.

Answer 1

for each segment of the trip,
s = vt + 1/2 at^2
so plug in your numbers.

Answer 2

Determine the total distance traveled by the car from its starting position to its stopping position.

Answer 3

Huh. I can see I need to get you started.
so, for the first part of the trip,
s = 0 + 5.75/2 * 6.76^2 = 131.38 m
Now do the 2nd part and the 3rd part, and add them up.

Answer 4

I got the second part, 131.38 and 284.53. How do I get the time for the 3rd part?