# In the first second of uniformly accelerated movement, the body travels a path of 1 m, and in the second – 2 m.

**In the first second of uniformly accelerated movement, the body travels a path of 1 m, and in the second – 2 m. Determine the path traversed by the body in the first three seconds of movement.**

t1 = 1 s.

S1 = 1 m.

t2 = 2 s.

S2ʹ = 2 m.

t3 = 3 s.

S3 -?

With uniformly accelerated motion, the path of the body is expressed by the formula: S3 = V0 * t3 + a * t32 / 2.

Let us express the path for the first and second second of the movement: S1 = V0 * t1 + a * t12 / 2, S2ʹ = S2 – S1.

S2 = V0 * t2 + a * t22 / 2.

S2ʹ = V0 * t2 + a * t22 / 2 – V0 * t1 + a * t12 / 2.

2 = V0 * 2 + a * (2) 2/2 – V0 * 1 – a * (1) 2/2 = V0 + 3 * a / 2.

a = 2 * (2 – 1 * V0) / 3.

S1 = V0 * t1 + (2 – 1 * V0) * t12 / 3 = V0 * 1 + (2 – 1 * V0) * (1) 2/3 = 1.

2 * V0 = 1.

The initial velocity of the body is V0 = 0.5 m / s.

We find the acceleration of the body by the formula: a = 2 * (2 – 1 * 0.5) / 3 = 1 m / s2.

S3 = 0.5 m / s * 3 s + 1 m / s2 * (3 s) 2/2 = 6 m.

Answer: in the first three seconds, the body passed the path S3 = 6 m.