The point of the beginning of rectilinear movement from a state of rest with a constant acceleration

The point of the beginning of rectilinear movement from a state of rest with a constant acceleration of 0.8 m / s ^ 2. 15 s after the beginning of the movement, the point began to move equally slowly and over the next 20 s, its speed decreased to 6 m / s. Determine: The path traversed by the point for the entire time of movement; the acceleration (deceleration) of the point in the section of equally slow motion.

V0 = 0 m / s.

a1 = 0.8 m / s2.

t1 = 15 s.

t2 = 20 s.

V2 = 6 m / s.

S -?

a2 -?

S = S1 + S2.

S1 = V0 * t1 + a1 * t1 ^ 2/2.

S1 = a1 * t1 ^ 2/2.

S1 = 0.8 m / s2 * (15 s) ^ 2/2 = 90 m.

V1 = V0 + a1 * t1.

V1 = a1 * t1.

V1 = 0.8 m / s2 * 15 s = 12 m / s.

S2 = V1 * t2 – a2 * t2 ^ 2/2.

a2 = (V1 – V2) / t2.

a2 = (12 m / s – 6 m / s) / 20 s = 0.3 m / s2.

S2 = V1 * t2 – a2 * t2 ^ 2/2.

S2 = 12 m / s * 20 s – 0.3 m / s2 * (20 s) ^ 2/2 = 180 m.

S = 90 m + 180 m = 270 m.

Answer: during braking, the acceleration was a2 = 0.3 m / s2, the path of the body was S = 270 m.