A particle moves along the x-axis and its position at time t is given by x(t) = $t^3$ - 6 $t^2$ + 9t, where x is in metres and t is in seconds. The motion starts at t = 0. The first instant after t = 0 at which the particle reverses its direction of motion is:
Options

Question

A particle moves along the x-axis and its position at time t is given by x(t) = $t^3$ - 6 $t^2$ + 9t, where x is in metres and t is in seconds. The motion starts at t = 0. The first instant after t = 0 at which the particle reverses its direction of motion is:

Options

G1 Physics · Accepted Answer

Velocity v = dx/dt = 3t^2 - 12t + 9 = 3(t - 1)(t - 3). For 0 < t < 1, v is positive, while for 1 < t < 3, v is negative. Therefore the velocity changes sign at t = 1 s, so the particle first reverses direction at 1 s. Hence, option A is correct.

A particle moves along the x-axis and its position at time t is given by x(t) = t^3 - 6t^2 + 9t, where x is in metres and t is in seconds. The motion starts at t = 0. The first instant after t = 0 at which the particle reverses its direction of motion is: