The body moves uniformly with no initial velocity. The path traversed by the body in one hundred and first second is 10 m more than the path traversed by the body in one hundredth second. What way did the body go in the first second?
The figure shows the sections that the body passed for each second of movement.
According to the condition of the problem, it is known that the path traveled by the body in the 101st second is 10 meters longer than the path covered in the 100th second. Therefore, taking into account the designations in the figure, we have:
S101 – S100 = 10 (m).
Similarly to the first problem, we express the speed of the body at the end of each segment in terms of acceleration:
ʋ0 = 0;
ʋ1 = a;
ʋ2 = 2a;
ʋ3 = 3a, you can notice the pattern and write the expression for ʋ99 and ʋ100:
ʋ99 = 99a;
ʋ100 = 100а.
It is necessary to find the path traveled by the body in the first second of movement, therefore, we use the formula S = ʋ0 * t + (a * t ^ 2) / 2, where ʋ0 is the speed at the beginning of the movement section, t is the time of movement in this section. And by condition, at each section of the movement t = 1 s.
Let us write this formula taking into account the designations in the figure for the path that the body has traveled in the hundredth and one hundred and first seconds of movement:
S101 = ʋ100 * 1 + (a * 1 ^ 2) / 2 = 100a + 0.5a = 100.5a;
S100 = ʋ99 * 1 + (a * 1 ^ 2) / 2 = 99a + 0.5a = 99.5a.
Substituting these expressions into the formula S101 – S100 = 10, we get:
100.5a – 99.5a = 10;
1a = 10;
a = 10 (m / s2).
Now you can determine the path traversed by the body in the first second of the movement:
S1 = ʋ0 * 1 + (a * 1 ^ 2) / 2 = 0 * 1 + (10 * 1 ^ 2) / 2 = 5 (m).
Answer: in the first second of movement, the body passed 5 meters.