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ques) A point mass starts moving in a straight line with constant acceleration. After time t o the acc changes its sign, remaining the same in magnitude. Dtermine the time t from the beginning of motion in which the point mass returns to its initial problems.

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11 years ago

```							Hi Tushar
Ans.  total time = 2t0
Apply v = u + at and find the velocity after time to(u = 0) and then after finding v , use this v to find the time t for the rest of the motion by jst changing the sign of acceleration and again apply v=u + at (final velocity = 0 ) .
add up the two times to get the total time.

```
11 years ago
```							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 t1²The 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 t1The total time T from initial point forward till back to initial point :          T  =  2 t1 + √2 t1 = (2 + √2) t1
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3 years ago
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