The position of a particle moving along the x-axis is given by x = 3.49t2 – 2.21t3, where x is in meters and t is in seconds. What is the position of the particle when it achieves its maximum speed in the positive x-direction?
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To find the position of the particle when it achieves its maximum speed in the positive x-direction, we need to analyze the given equation and determine the velocity of the particle first.
The velocity of a particle is the derivative of its position function with respect to time. So, let's find the velocity function by differentiating the given equation with respect to t.
x = 3.49t^2 - 2.21t^3
Let's differentiate both sides of the equation:
dx/dt = d/dt (3.49t^2 - 2.21t^3)
To find the derivative, we'll use the power rule, which states that the derivative of t^n is equal to n*t^(n-1).
dx/dt = 2 * 3.49t - 3 * 2.21t^2
Simplifying further:
dx/dt = 6.98t - 6.63t^2
Now we have the velocity function, which represents the velocity of the particle at any given time.
To find the maximum speed, we need to find the maximum value of the velocity function. To do that, we can find the value of t where the velocity is equal to zero. We'll set dx/dt = 0 and solve for t:
6.98t - 6.63t^2 = 0
Factoring out t:
t(6.98 - 6.63t) = 0
Setting each factor equal to zero:
t = 0
6.98 - 6.63t = 0
Solving the second equation for t:
6.63t = 6.98
t = 6.98 / 6.63
t ≈ 1.055 seconds
So, when t ≈ 1.055 seconds, the particle achieves its maximum speed in the positive x-direction.
Now that we have the time when the maximum speed occurs, we can substitute this value of t back into the position function to find the position of the particle at that time:
x = 3.49t^2 - 2.21t^3
Substituting t ≈ 1.055:
x = 3.49(1.055)^2 - 2.21(1.055)^3
Evaluating this expression will give us the final answer for the position of the particle when it achieves its maximum speed in the positive x-direction.