The body, moving uniformly accelerated from the state of rest, traveled in 12 s a distance of 450 m.

The body, moving uniformly accelerated from the state of rest, traveled in 12 s a distance of 450 m. At what distance from the initial position was it 4 s after the start of movement.

Given:

V0 = 0;

t = 12 s;

S = 450 m.

t4 = 4 s.

S4 =?

Decision:

1. Let’s use the formula for uniformly accelerated motion S = V0 * t + at ^ 2: 2.

Since the initial velocity is V0 = 0, then S = at ^ 2: 2.

2. Let us express the acceleration from the formula: a = 2S: t ^ 2;

a = 450 * 2: 12 ^ 2 = 900: 144 = 6.25 m / s.

3. Let’s calculate the distance covered in 4 seconds:

S4 = a * t4 ^ 2: 2 = 6.25 * 4 ^ 2: 2 = 6.25 * 16: 2 = 100: 2 = 50 m.

Answer: 4 minutes after the start of the movement, the body was at a distance of 50 meters from the initial position.