A particle starts from rest at the origin and moves along the x -axis. Its acceleration varies with time t as a = 12t + 6 (where a is in m/s ^2 and t is in seconds). The ratio of its displacement to its velocity at t = 2 s is:

Step-by-step solution

Given acceleration a = 12t + 6 Velocity v = a dt = (12t + 6) dt = 6t^2 + 6t + C₁ Since the particle starts from rest, v = 0 at t = 0 , which gives C₁ = 0 . v = 6t^2 + 6t Displacement x = v dt = (6t^2 + 6t) dt = 2t^3 + 3t^2 + C₂ Since the particle starts from the origin, x = 0 at t = 0 , which gives C₂ = 0 . x = 2t^3 + 3t^2 At t = 2 s : v = 6(2)^2 + 6(2) = 24 + 12 = 36 m/s x = 2(2)^3 + 3(2)^2 = 16 + 12 = 28 m Ratio of displacement to velocity is x v = 28 36 = 7 9 Answer: 7 : 9