The velocity of a particle moving along the x axis is given by vx = a t - b t3 for t > 0, where a = 27 m/s 2, b = 3.1 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 in units of m/s 2.

-54 m/s²

Explanation:

Acceleration is defined as the change in velocity of a body with respect to time. Mathematically,

Acceleration A = change in velocity/time

A = dv/dt

Given Vx = at - bt³

The time at which the particle reaches its maximum displacement is at when vx = 0.

0 = at-bt³

t (a-bt²) = 0

a-bt² = 0

a = bt²

t² = a/b ... (1)

A = dvx/dt = a - 3bt² (by differentiating)

Acceleration = a - 3bt² ... (2)

Substituting t² = a/b into equation 2 will give;

Acceleration = a - 3b (a/b)

Acceleration = a-3a

Acceleration = - 2a

Substituting the value of a = 27m/s into the resulting equation of acceleration gives;

Acceleration = - 2 (27)

Acceleration = - 54m/s²

Therefore at maximum displacement in the positive x direction, the acceleration of the particle will be - 54m/s²