Question

A point mass starts moving in a straight line with constant acceleration 'a'. At a time t after the beginning of motion, the acceleration changes sign, without change in magnitude. Determine the time t₀ from the beginning of the motion in which the point mass returns to the initial position. ((2+√2)t)

Answer 1

let intial velocity is u = 0 , acceleration is a.
time taken to cover distance S.
use formula, S = ut + 1/2 at²
S = 1/2 at² ----(1)
and final velocity , v = u + at = 0 + at
e.g., v = at -----(2)

now, sign of acceleration changes . but for some distance body move same direction untill it becomes rest.
so, final velocity in this case , v = 0
initial velocity , u = at [ because final velocity of 1st case is considered initial velocity of 2nd case.]
use formula, v = u + at
0 = at + (-a)t' => t' = t
so, S' = ut' + 1/2(-a)t'²
S'= at² -1/2at² = 1/2at² ------(3)

now, body moves opposite direction and comes to intial point.
so, initial velocity in this case, u = 0
and acceleration = a
use formula, S" = ut" + 1/2at"²
S + S' = 1/2at² + 1/2at² = 0 1/2at"²
at² = 1/2at"² => t" = √2t

so, total time = t + t' + t"
= t + t + √2t = (2 + √2)t