What is the acceleration of an asteroid at rest during its movement across a screen, according to its motion equation x=6.5t-2.3t^3?

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What is the acceleration of an asteroid at rest during its movement across a screen, according to its motion equation x=6.5t-2.3t^3?

To determine the acceleration of the asteroid using the motion equation \( x = 6.5t - 2.3t^3 \), we need to first find the velocity by taking the derivative of the position equation with respect to time \( t \). The position function is given as \( x(t) = 6.5t - 2.3t^3 \). Differentiating this with respect to \( t \) yields the velocity: \[ v(t) = \frac{dx}{dt} = 6.5 - 6.9t^2 \] Next, to find the acceleration, we take the derivative of the velocity function: \[ a(t) = \frac{dv}{dt} = -13.8t \] Since the acceleration depends on time through the variable \( t \), we will evaluate it at a specific point in time. If the asteroid is at rest, it suggests we can consider the situation when \( t = 0 \) because at this point the asteroid is not in motion. Evaluating the acceleration at \( t = 0 \): \[ a(0) = -13.8 \times 0 = 0 \]

To determine the acceleration of the asteroid using the motion equation ( x = 6.5t - 2.3t^3 ), we need to first find the velocity by taking the derivative of the position equation with respect to time ( t ).

The position function is given as ( x(t) = 6.5t - 2.3t^3 ). Differentiating this with respect to ( t ) yields the velocity:

[

v(t) = \frac{dx}{dt} = 6.5 - 6.9t^2

]

Next, to find the acceleration, we take the derivative of the velocity function:

[

a(t) = \frac{dv}{dt} = -13.8t

]

Since the acceleration depends on time through the variable ( t ), we will evaluate it at a specific point in time. If the asteroid is at rest, it suggests we can consider the situation when ( t = 0 ) because at this point the asteroid is not in motion.

Evaluating the acceleration at ( t = 0 ):

[

a(0) = -13.8 \times 0 = 0

]

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