Moving with an acceleration of 5m / s2, the body on the path of 600m increased its speed by 4 times.

Moving with an acceleration of 5m / s2, the body on the path of 600m increased its speed by 4 times. The initial velocity of the body is.

Given:

a = 5 m / s ^ 2 – body acceleration;

S = 600 meters – the section traversed by the body;

v1 = 4 * v0 – in this area, the body has increased its speed by 4 times.

It is required to determine the initial velocity of the body v0 (m / s).

Let’s find the time of body movement:

t = (v1 – v0) / a = (4 * v0 – v0) / a = 3 * v0 / a.

Then the path traversed by the body is:

S = v0 * t + a * t ^ 2/2 = v0 * 3 * v0 / a + a * 9 * v0 ^ 2/2 * a =

= 3 * v0 ^ 2 / a + 9 * v0 ^ 2/2 * a = 15 * v0 ^ 2/2 * a, hence:

v0 = (2 * a * S / 15) ^ 0.5 = (2 * 600 * 5/15) ^ 0.5 = (6000/15) ^ 0.5 = 400 ^ 0.5 = 20 m / s …

Answer: the initial velocity of the body is 20 m / s.