 # ques) A point mass starts moving in a straight line with constant acceleration. After time to 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.

14 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.

6 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