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 $t_°$
(t not)from the beginning of the motion in which the point of mass returns to the initial position?

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Answer 1

Equations of motion are:

   s = u t + 1/2  a t²  ,

         u = 0  and at time t1, when the acceleration changes, distance travelled:

     s = 1/2 a t1²

     v at t = t1 =  u + a t  = a t1

 

Now the acceleration is changed to -a.  Then the particle continues in the same direction until the velocity becomes zero.  Then the particle changes the direction and starts accelerating and passes over the point of start.

     u = a t1      v = 0    acceleration = -a

       v = u + a t

  => 0 = a t1 - a t   =>    t = t1    it takes  t1 more time to stop and reverse direction.

  The distance traveled/displacement in this time:

      s = u t + 1/2 a t²

   => s = a t1 * t1 - 1/2 a  t1² = 1/2 a t1²

The total displacement from the initial point :  1/2 a t1² + 1/2 a t1² = a t1²

   now,  acceleration = -a    u = 0       s = - a t1²   in the negative direction

     s = u t + 1/2  a  t²

   => - a t1² = 0 - 1/2 a t²

   =>   t = √2 t1

The total time T from initial point forward till back to initial point :

          T  =  2 t1 + √2 t1 = (2 + √2) t1

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Answer 2

 s = u t + 1/2  a t²  ,

        u = 0  and at time t1, when the acceleration changes, distance travelled:

    s = 1/2 a t1²

    v at t = t1 =  u + a t  = a t1

 

Now the acceleration is changed to -a.  Then the particle continues in the same direction until the velocity becomes zero.  Then the particle changes the direction and starts accelerating and passes over the point of start.

    u = a t1      v = 0    acceleration = -a

      v = u + a t

 => 0 = a t1 - a t   =>    t = t1    it takes  t1 more time to stop and reverse direction.

 The distance traveled/displacement in this time:

     s = u t + 1/2 a t²

  => s = a t1 * t1 - 1/2 a  t1² = 1/2 a t1²

The total displacement from the initial point :  1/2 a t1² + 1/2 a t1² = a t1²

  now,  acceleration = -a    u = 0       s = - a t1²   in the negative direction

    s = u t + 1/2  a  t²

  => - a t1² = 0 - 1/2 a t²

  =>   t = √2 t1

The total time T from initial point forward till back to initial point :

         T  =  2 t1 + √2 t1 = (2 + √2) t1

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