A particle starts from rest and has an acceleration of 2 ~m / s ^2 for 10 sec . After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4 ~m / s ^2 and comes back to rest. The total distance covered by the particle is

Step-by-step solution

Initial velocity (u)=0 , Acceleration (a₁ )=2 ~m / s ^2 and time during acceleration (t₁ )=10 sec . Time during constant velocity (t₂ )=30 sec and retardation (a₂ )=-4 ~m / s ^2 (-ve sign due to retardation). Distance covered by the particle during acceleration, gathered s₁=u t₁+ 1 2 a₁ t₁^2=(0 10)+ 1 2 2 (10)^2 =100 ~m gathered And velocity of the particle at the end of acceleration, v=u+a₁ t₁=0+(2 10)=20 ~m / s . Therefore distance covered by the particle during constant velocity (s₂ )=v t₂ =20 30=600 ~m Relation