Question

An object moves along the x-axis and its position is given by the function ()=1/6 ⁴−13/6 ³+3t²+2 . Find the time(s), , at which the acceleration is 0.

Answer 1

To find the times at which the acceleration is zero, one must calculate the second derivative of the given position function, set it equal to zero, and solve for t using the quadratic formula, excluding any negative time values.

Step-by-step explanation:

The question is asking to find the time(s) at which the acceleration of an object moving along the x-axis is 0. To answer this, we need to find the second derivative of the position function, set it equal to 0, and solve for t.

Given the position function as x(t) = (1/6)t´ - (13/6)t³ + 3t² + 2, we first find the velocity function by taking the first derivative:

v(t) = dx/dt = (4/6)t³ - (39/6)t² + 6t

Next, we find the acceleration function by taking the second derivative of the position function, or the first derivative of the velocity function:

a(t) = dv/dt = (12/6)t² - (78/6)t + 6

Simplify this to get a(t) = 2t² - 13t + 6. To find the time(s) when acceleration is 0, we solve the equation 2t² - 13t + 6 = 0:

Using the quadratic formula, t = [13 ± sqrt(13² - 4*2*6)]/(2*2)

Calculate the discriminant and simplify to find the values of t that make the acceleration equal to 0. Exclude any negative values as they would represent times before the motion starts, which are not physically meaningful in this context.