# The material point moves in a straight line with an acceleration of 5 m / s2.

**The material point moves in a straight line with an acceleration of 5 m / s2. Determine how much the path traveled in the n-th second, there will be more path traveled in the previous second. take V0 = 0**

If we consider not the 1st and 2nd second of the movement, then the initial speed will not be equal to 0, since the point has already been on the way for some time.

v = v0 + at;

v0 (for the (n-1) th second) = 0 + a * (n – 2) = 5 * (n – 2) = 5n – 10;

v0 (for the nth second) = 0 + 5 * (n – 1) = 5n – 5;

S = v0 * t + at ^ 2/2;

t = 1 s;

For the (n-1) th and n-th second, the distance traveled will be, respectively:

S1 = (5n – 10) * 1 + 5 * 12/2 = 5n – 7.5;

S2 = (5n – 5) * 1 + 5 * 12/2 = 5n – 2.5;

S2 – S1 = (5n – 2.5) – (5n – 7.5) = 5 m.

Answer: in each subsequent second, the point will pass 5 meters more than in the previous one.