The car from a state of rest accelerates with an acceleration of 1 m / s2, then moves at a speed of 54 km / h

The car from a state of rest accelerates with an acceleration of 1 m / s2, then moves at a speed of 54 km / h, and then decelerates with an acceleration of -3 m / s2 to a complete stop. What is the movement of the car? The uniform movement time is 20 s.

Divide the path into parts
1). Moves from rest with acceleration
v0 = 0 m / s, a = 1 m / s ^ 2
S = (v ^ 2 – v0 ^ 2) / 2a
The car accelerated to 54 km / h, i.e. 15 m / s
v = 15 m / s
S = (15 ^ 2 – 0 ^ 2) / 2 * 1
S = (225-0) / 2
S = 112.5 (m)
2) Moves evenly
Speed 15 m / s, time 20s
S = v * t
S = 15 * 20
S = 300 m
3) Brakes with an acceleration of -3 m / s ^ 2 (minus as it slows down) to a complete stop
This means that in this section the initial speed is v0 = 15m / s, the final speed is v = 0 m / s
S = (v ^ 2 – v0 ^ 2) / 2a
S = (0 – 15 ^ 2) / 2 * (- 3)
S = – 225 / (- 6)
S = 37.5 (m)
It remains to fold the full path in all sections
S = 112.5 m + 300 m + 37.5 m = 450 m
Answer: 450 m.