The position of a particle moving along the x axis is given by x=3.0t^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

If x is the displacement, note that

v (velocity) = dx/dt, and
a (acceleration) = dv/dt
Therefore, we get the derivative of the given equation:
dx/dt = 6t - 3t^2 = velocity
Note that this is just the equation for velocity, and we need to maximize it. Thus, we differentiate it again:
dv/dt = 6 - 6t = acceleration
We equate this to zero and solve for t:
0 = 6 - 6t
6t = 6
t = 1 s
At t = 1, the velocity is maximum, and the displacement is
x = 3.0t^2 - 1.0t^3
x = 3*(1^2) - (1^3)
x = 3 - 1
x = 2 m

Hope this helps :3

Well, when a particle achieves its maximum speed, it means that it is at the point where its acceleration changes direction. This happens when velocity is at its maximum. So, let's first find the velocity.

The velocity of the particle is given by the derivative of the position equation with respect to time.

v = dx/dt = d(3.0t^2 - 1.0t^3)/dt

v = 6.0t - 3.0t^2

To find the maximum speed, we need to find when the velocity is at its maximum. To do that, let's take the derivative of the velocity equation with respect to time.

a = dv/dt = d(6.0t - 3.0t^2)/dt

a = 6.0 - 6.0t

To find when the acceleration changes direction, we need to set the acceleration equal to zero and solve for t.

6.0 - 6.0t = 0
6.0t = 6.0
t = 1.0

So, when t = 1.0 second, the particle achieves its maximum speed in the positive x direction.

Now, let's substitute this value of t into the position equation to find the position of the particle at that time.

x = 3.0t^2 - 1.0t^3
x = 3.0(1.0)^2 - 1.0(1.0)^3
x = 3.0 - 1.0
x = 2.0 meters

Therefore, the position of the particle when it achieves its maximum speed in the positive x direction is 2.0 meters.

To find the position of the particle when it achieves its maximum speed in the positive x direction, we need to find the time at which the particle has its maximum speed.

The speed of the particle is given by the derivative of the position function with respect to time:

v(t) = d(x)/dt = d/dt (3.0t^2 - 1.0t^3)

Let's calculate the derivative:

v(t) = 6.0t - 3.0t^2

To find the maximum speed, we need to find the time at which v(t) = 0. We solve the equation:

6.0t - 3.0t^2 = 0

Factorizing it, we get:

t(6.0 - 3.0t) = 0

So we have two possibilities:

1. t = 0
2. 6.0 - 3.0t = 0

Solving the second equation, we find:

6.0 - 3.0t = 0
-3.0t = -6.0
t = 2.0

Now that we have the time at which the particle achieves maximum speed, we substitute this value into the position function to find the position:

x = 3.0t^2 - 1.0t^3
x = 3.0(2.0)^2 - 1.0(2.0)^3
x = 3.0(4.0) - 1.0(8.0)
x = 12.0 - 8.0
x = 4.0

Therefore, the position of the particle when it achieves its maximum speed in the positive x direction is 4.0 meters.

To find the position of the particle when it achieves its maximum speed in the positive x direction, we need to first determine the velocity function and then find the time at which the velocity is maximum.

The velocity function is the derivative of the position function with respect to time, which gives us:

v = dx/dt

To find v, let's take the derivative of the position function x with respect to t:

x = 3.0t^2 - 1.0t^3

Taking the derivative:

v = dx/dt = d(3.0t^2 - 1.0t^3)/dt

v = 6.0t - 3.0t^2

Next, we need to find the time (t) at which the velocity (v) is maximum. To do this, we set the derivative of the velocity function equal to zero and solve for t:

dv/dt = 6.0 - 6.0t = 0

Solving for t:

6.0 - 6.0t = 0
6.0t = 6.0
t = 1.0

So at t = 1.0 second, the velocity is maximum in the positive x direction.

To find the position of the particle at this time, we substitute t = 1.0 second into the position function:

x = 3.0t^2 - 1.0t^3
x = 3.0(1.0)^2 - 1.0(1.0)^3
x = 3.0 - 1.0
x = 2.0 meters

Therefore, the position of the particle when it achieves its maximum speed in the positive x direction is 2.0 meters.