# Find the path traveled by the body, the speed of which changes according to the law v = 2–2t,

**Find the path traveled by the body, the speed of which changes according to the law v = 2–2t, 4 s from the beginning of the movement.**

v (t) = 2 – 2 * t.

t = 4 s.

S -?

When a body moves with a constant acceleration a, then its velocity v (t) changes according to the law: v (t) = v0 + a * t, where v0 is the initial velocity of the body, t is the time of motion.

For our case, v (t) = 2 – 2 * t, at the initial moment the body had a velocity v0 = 2 m / s, it moves with an acceleration a = 2 m / s2, which is directed in the opposite direction of the body’s motion, it is decelerated to full stops and then starts to move in the opposite direction.

With uniformly accelerated motion S, the path is expressed by the formula: S = S1 + S2.

2 – 2 * t1 = 0.

t1 = 1 s – deceleration time.

t2 = t – t1 = 4 s – 1 s = 3 s – acceleration time.

S1 = v0 * t1 – a * t1 ^ 2/2.

S1 = 2 m / s * 1 s – 2 m / s2 * (1 s) ^ 2/2 = 1 m.

S2 = a * t2 ^ 2/2.

S1 = 2 m / s2 * (3 s) ^ 2/2 = 9 m.

S = 1 m + 9 m = 10 m.

Answer: the path of the body was S = 10 m.