A point mass starts moving in straight line with constant acceleration a from rest at t=0.At time t=2s ,the acceleration changes the sign,remaining the same in magnitude .The mass returns to the initial position at time t=t0 after start of motion .Here, t0 is=? Ans is (4+2root2)s but how?

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Solution

Given initial velocity of particle is u =0 , acceleration = a, from A to b and -a from B to A
time taken by particle to go from A to B (t₁) =2 second , let t₂ be the time in going from B to A
now t₀ = t₁+ t₂
firstly writing equation from A to B
final velocity at time t = 2 second
= v= u+at₁ = 0+2a =2a .......1>
s= ut₁+1/2at₁² = 0+1/2 a(2)²
s = 2a ..............................2>
now writing the equation fromB to A
s = ut₂-1/2at₂²= 2at₂-1/2 at₂²( v=u since at B the body reverse its direction and hence final velocity of A to B i
initial velocity from B to A)
-2a = 2at₂-1/2 at₂²(putting value of s from eq 2 since AB = BA since the direction of displacement is opposite to initial)
$- 2 = 2 t_{2} - \frac{1}{2} t^{2} (c a n c e l l i n g a t o b o t h s i d e s) - 2 = \frac{4 t_{2} - t^{2}}{2} - 4 = 4 t_{2} - t^{2} t^{2} - 4 t_{2} - 4 = 0 (q u a d r a t i c e q u a t i o n) t_{2} = \frac{- (- 4) + \sqrt{16 + 16}}{2} (n e g l e c t i n g n e g e t i v e p a r t \sin c e t i m e c a n n o t b e n e g e t i v e) t_{2} = \frac{4 + 4 \sqrt{2}}{2} = 2 + 2 \sqrt{2}$
now
t₀ = t₁ + t₂ = 2+2+2 $\sqrt{2}$
= 4+2 $\sqrt{2}$ .

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