The velocity of a particle moving along the x axis is given for 1> 0 by v: = (32.0t - 2.00t^3) m/s,

where t is in s. What is the acceleration of the particle when (after t = 0) it achieves its maximum

displacement in the positive x direction?

-64.0 m/s²

Explanation:

v = 32.0 t - 2.00 t³

Acceleration is the derivative of velocity.

a = dv/dt

a = 32.0 - 6.00 t²

At maximum displacement, v = 0 and a < 0.

v = 0

32.0 t - 2.00 t³ = 0

2.00 t (16.0 - t²) = 0

t = 0 or 4.00

When t = 0 s, a = 32.0 m/s².

When t = 4.00 s, a = - 64.0 m/s².

The maximum displacement is at t = 4.00 s. Therefore, the acceleration at that moment is - 64.0 m/s².

Answer v = 32.0 t - 2.00 t³

Acceleration is the derivative of velocity.

a = dv/dt

a = 32.0 - 6.00 t²

At maximum displacement, v = 0 and a < 0.

v = 0

32.0 t - 2.00 t³ = 0

2.00 t (16.0 - t²) = 0

t = 0 or 4.00

When t = 0 s, a = 32.0 m/s².

When t = 4.00 s, a = - 64.0 m/s².