# The body moves uniformly with no initial velocity. In the third second

**The body moves uniformly with no initial velocity. In the third second, the body traveled 15 m. What path did the body cover in the second second?**

The figure shows three sections, each of which the body passed in a second.

In all areas, the body moved with the same acceleration, since the movement is uniformly accelerated. Acceleration is determined by the formula a = (ʋ – ʋ0) / t. Where ʋ0 is the speed of the body at the beginning of the movement, ʋ is the speed at the end of the movement. Let’s write down the expression of acceleration for each section, taking into account the designations in the figure. We get:

a = (ʋ1 – ʋ0) / 1 = ʋ1 – ʋ0, since the time is 1 sec .;

a = (ʋ2 – ʋ1) / 1 = ʋ2 – ʋ1;

a = (ʋ3 – ʋ2) / 1 = ʋ3 – ʋ2.

From these equalities we find the expression for the speed in terms of the acceleration of the body:

ʋ1 = a;

ʋ2 = a + ʋ1 = a + a = 2a;

ʋ3 = a + 2a = 3a.

Recall that the path that the body has traveled over a certain time interval can be found by the formula:

S = ʋ0 * t + (a * t ^ 2) / 2. Where ʋ0 is the speed at the beginning of the section of movement, t is the time of movement in this section. And by condition t = 1 s.

Substitute in this formula the designations of quantities according to the figure:

S3 = ʋ2 * t + (a * t ^ 2) / 2 = 2a * 1 + (a * 1 ^ 2) / 2 = 2a + a / 2 = 2a + 0.5a = 2.5a.

According to the condition of the problem, S3 = 15 m, which means 2.5a = 15 m; that is, a = 6 (m / s2).

Now you can find the path that the body has traveled in the second second of movement using the formula:

S2 = ʋ1 * t + (a * t ^ 2) / 2 = a * 1 + (a * 1 ^ 2) / 2 = a + a / 2 = 1.5 a = 1.5 * 6 = 9 (m ).

Answer: in the second second of movement, the body will travel 9 meters.