The body, moving uniformly accelerated, passes two consecutive identical sections of the path of 15 m

The body, moving uniformly accelerated, passes two consecutive identical sections of the path of 15 m each, respectively, in 2 s and 1 s. Determine the initial speed of the body.

Given:

L1 = L2 = 15 meters – lengths of successive track sections;

t1 = 2 seconds – the time during which the body passes the first leg of the path;

t2 = 1 second – the time during which the body passes the second section of the path.

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

Let the acceleration of the body be equal to a. Then we have the following equations:

L1 = v0 * t1 + a * t1 ^ 2/2 (1);

L2 = v * t ^ 2 + a * t2 ^ 2/2 (2);

Taking into account that v = v0 + a * t1, we get:

L2 = (v0 + a * t1) * t ^ 2 + a * t2 ^ 2/2 = (v0 + a * 2) * 1 + a * 1 ^ 2/2 = v0 + 2 * a + a / 2 = v0 + 5 * a / 2 (3).

From equation (1) we find that the acceleration of the body is equal to:

a = 2 * (L1 – v0 * t1) / t1 ^ 2 = 2 * (15 – v0 * 2) / 2 ^ 2 = (15 – 2 * v0) / 2 = 7.5 – v0.

Substituting the found acceleration value into equation (3), we obtain:

L2 = v0 + 5 * a / 2;

15 = v0 + 5 * (7.5 – v0) / 2 = v0 + 18.75 – 2.5 * v0 = 18.75 – 1.5 * v0, from here we find that:

v0 = (18.75 – 15) / 1.5 = 2.5 m / s.

Answer: the initial velocity of the body was 2.5 m / s.