Answer to Question #220822 in Mechanics | Relativity for jafar

Question

the position of a particle moving along the x axis is given by x =t^2-2t^3-6,where x is in meters and t in second. What is the position of the particle when it achieves its maximum speed in the positive x direction?

Answer 1

Let us determine the moment of the maximum positive speed.

$v_x = x' = (t^2 - 2t^3-6)' = 2t - 6t^2.$

The maximum velocity corresponds to $v_x' = 0,$ and $v_x' = 2 - 12t.$ Therefore, $v_x' = 0$ when $t = \dfrac16.$ The velocity is $v_x = 2\cdot\dfrac16 - 6\cdot\left(\dfrac16\right)^2 = \dfrac16$ and the position is $x= \left(\dfrac16\right)^2 - 2\left(\dfrac16\right)^3 - 6 \approx -5.98.$