A particle is moving in a straight line. and passes through a point o with a velocity of 6 ms-1.The particle moves with a constant retardation of 2ms-2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O?

Let the particle moves toward right with velocity 6 m/s. Due to retardation, after time t1, its velocity becomes zero. From v=u−at⇒0=6−2×t1⇒ t1=3secBut retardation works on it for 4 sec. It means after reaching point A, direction of motion gets reversed and acceleration works on the particle for next one second.SOA=ut1−12at12=6×3−12(2)(3)2=18−9=9mSAB=12×2×(1)2=1m∴ SBC=S0A−SAB=9−1=8mNow velocity of the particle at point B in return journeyv=0+2×1=2m/sIn return journey from B to C, particle moves with constant velocity 2 m/s to cover the distance 8 m. Time taken = Distance  Velocity =82=4sTotal time taken by particle to return at point 0 isT=t0A+tAB+tBC=3+1+4=8s.