A body starting from rest at t=0 moves along a straight line with a constant acceleration. At t=2 ~s , the body reverses its direction keeping the acceleration same. The body returns to the initial position at t=t₀ , then t₀ is

Step-by-step solution

According to the question, From first equation of the motion, v₁=u+a t₁ v₁=2 a Firstly, body decelerate with acceleration to the point C and then reverse it's direction and accelerate with acceleration a to the point A . Therefore for distance B C , from first equation of the motion, or v₂=v₁-a t₂ 0=2 a-a t₂ t₂=2 ~s Hence, total time taken by body to covered distance A C, t=2+2=4 ~s From second equation one motion, array rlrl & s₁ & =A B=u t₁+ 1 2 a t₁^2=0+ 1 2 a 2^2 & & s₁ & =2 a & & s₁ & =s₂=2 a & A C & =s₁+s₂=4