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.