Distance/time Physics Problem?

Question

Problem: A car traveling 55 km/h slows down at a constant 0.50 m/s^2. Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.

v(initial) = velocity(initial) = 55km/h = 15.28m/s

v = velocity

a = acceleration = -0.50m/s^2

For (a), I used the equation v^2 = v(initial)^2 + 2a(x - x(initial)) to get the resulting displacement of 230 meters.

For (b), I found the time using the equation v = v(initial) + at, which resulted in 31 seconds.

However, I really have no idea how to approach (c). Any help would be greatly appreciated.

Answer 1

"first and fifth seconds" is a very ambiguous phrase.

Assuming it means "from t = 0s to t = 1s" and "t = 4s to t = 5s",

d = Vo*t + ½a*t²

Solve this eq for the above intervals, remembering that 'a' is a negative value