Question

The velocity of a particle moving along the x axis is given by vx = a t − b t3 for t > 0 , where a = 26 m/s 2 , b = 1.4 m/s 4 , and t is in s. What is the acceleration of the particle when it achieves its maximum displacement in the positive x direction?

Answer 1

-52 m/s²

Step-by-step explanation:

`a=26 m/s^2`

`b=1.4 m/s^4`

Velocity is given by

`v=at-bt^3`

Acceleration is given by

`a_x=(dv)/(dt)\\\Rightarrow a_x=(d)/(dt)at-bt^3\\\Rightarrow a_x=a-3bt^2`

For maximum displacement we have to equate v = 0

`0=at-bt^3\\\Rightarrow t=\sqrt{(a)/(b)}\\\Rightarrow t=\sqrt{(26)/(1.4)`

Substituting in the acceleration equation

`a_x=26-3* 1.4* \sqrt{(26)/(1.4)}^2\\\Rightarrow a_x=26-3* 1.4* (26)/(1.4)\\\Rightarrow a_x=-52\ m/s^2`

The acceleration when maximum displacement is achieved is -52 m/s²