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

]