A point mass starts moving in a straight line with constant acceleration. After time $t_{0}$ the acceleration changes its sign, remaining the same in magnitude. Determine the time T from the beginning of motion in which the point mass returns to the initial position.

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Solution

$v e l o c i t y a t t_{0} = a t_{0} d i s p l a c e m e n t = \frac{1}{2} {a t_{0}}^{2} t o r e t u r n t o t h e s a m e p o i n t d i s p l a c e m e n t i s - \frac{1}{2} {a t_{0}}^{2}, a = - a, a n d u = a t_{0} - \frac{1}{2} {a t_{0}}^{2} = a t_{0} t - \frac{1}{2} a t^{2} \Rightarrow t = t_{0} (1 \pm \sqrt{2}) \Rightarrow t = t_{0} (1 + \sqrt{2}) t o t a l t i m e f r o m t h e b e g i n n i n g o f m o t i o n = t + t_{0} = t_{0} (2 + \sqrt{2})$

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A point mass starts moving in a straight line with constant acceleration. After time t0 the acceleration changes its sign, remaining the same in magnitude. Determine the time T from the beginning of motion in which the point mass returns to the initial position.

velocityatt0=at0displacement=12at02toreturntothesamepointdisplacementis−12at02,a=−a,andu=at0−12at02=at0t−12at2⇒t=t0(1±√2)⇒t=t0(1+√2)totaltimefromthebeginningofmotion=t+t0=t0(2+√2)

A point mass starts moving in a straight line with constant acceleration. After time $t_{0}$ the acceleration changes its sign, remaining the same in magnitude. Determine the time T from the beginning of motion in which the point mass returns to the initial position.