A particle moves along the x-axis such that its position at time t is given by x(t) = t^3 3 - 2t^2 + 3t . The ratio of the total distance covered to the magnitude of the net displacement of the particle in the time interval t = 0 to t = 4 s is:

Step-by-step solution

The position of the particle is given by x(t) = t^3 3 - 2t^2 + 3t . First, find the velocity to determine the turning points: v(t) = dx dt = t^2 - 4t + 3 Setting v(t) = 0 gives t^2 - 4t + 3 = 0 (t - 1)(t - 3) = 0 . The particle turns at t = 1 s and t = 3 s. Now, evaluate the position at the boundaries of the time interval and at the turning points: At t = 0 : x(0) = 0 At t = 1 : x(1) = 1 3 - 2 + 3 = 4 3 At t = 3 : x(3) = 27 3 - 2(9) + 3(3) = 9 - 18 + 9 = 0 At t = 4 : x(4) = 64 3 - 2(16) + 3(4) = 64 3 - 32 + 12 = 64