Question

The position of a particle moving along the x axis is given by x = 6.0t 2 - 1.0t 3, where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

Answer 1

Answer:

The position of the particle when it achieves its maximum speed in the positive x direction is 20 meters

Explanation:

The position of the particle is given by the equation

$x=6t^2-t^3$

The speed (v) of the particle is calculated as:

$\begin{gathered} v=\frac{dx}{dt} \\ v=12t-3t^2 \end{gathered}$

The acceleration of the particle is:

$\begin{gathered} a=\frac{dv}{dt} \\ a=12-6t \end{gathered}$

At maximum speed, the acceleration of the particle is zero

That is, a = 0

a = 12 - 6t

0 = 12 - 6t

6t = 12

t = 12/6

t = 2

This means that the speed of the particle is maximum at the time, t = 2

Substitute t = 2 into x = 6t² - t³

x = 6(2²) - (2²)

x = 6(4) - 4

x = 24 - 4

x = 20 meters

The position of the particle when it achieves its maximum speed in the positive x direction is 20 meters