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A particle moves along the
x-axis so that at time
t >= 0 its position is given by
x(t)=t^(3)-3t^(2)-9t. Determine all values of
t when the particle is at rest.
Answer:
t=

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=t^{3}-3 t^{2}-9 t$ . Determine all values of $t$ when the particle is at rest. $\newline$ Answer: $t=$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle moves along the

x

-axis so that at time

t \geq 0

its position is given by

x(t)=t^{3}-3 t^{2}-9 t

. Determine all values of

t

when the particle is at rest.

\newline

Answer:

t=

Find Velocity Function: To find when the particle is at rest, we need to determine when its velocity is zero. The velocity of the particle is the derivative of its position function $x(t)$ . Let's find the derivative of $x(t) = t^3 - 3t^2 - 9t$ . $v(t) = \frac{dx}{dt} = \frac{d}{dt} (t^3 - 3t^2 - 9t)$ $v(t) = 3t^2 - 6t - 9$
Solve for Zero Velocity: Now we need to solve for $t$ when the velocity $v(t)$ is zero. $\newline$ $0 = 3t^2 - 6t - 9$ $\newline$ To solve this quadratic equation, we can either factor it, complete the square, or use the quadratic formula. Let's try to factor it first.
Apply Quadratic Formula: We notice that the quadratic equation $3t^2 - 6t - 9$ does not factor easily. Therefore, we will use the quadratic formula to find the values of $t$ . The quadratic formula is given by $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 3$ , $b = -6$ , and $c = -9$ .
Calculate Discriminant: Let's calculate the discriminant $b^2 - 4ac$ first. $\newline$ Discriminant = $(-6)^2 - 4(3)(-9)$ $\newline$ Discriminant = $36 + 108$ $\newline$ Discriminant = ${144}$
Find Real Solutions: Since the discriminant is positive, we have two real solutions for $t$ . Now we will apply the quadratic formula. $t = \frac{-(-6) \pm \sqrt{144}}{2 \times 3}$ $t = \frac{6 \pm 12}{6}$
Final Valid Solution: We have two possible solutions for $t$ : $t = \frac{{6 + 12}}{{6}}$ and $t = \frac{{6 - 12}}{{6}}$ $t = \frac{{18}}{{6}}$ and $t = \frac{{-6}}{{6}}$ $t = 3$ and $t = -1$
Final Valid Solution: We have two possible solutions for $t$ : $t = (6 + 12) / 6$ and $t = (6 - 12) / 6$ $t = 18 / 6$ and $t = -6 / 6$ $t = 3$ and $t = -1$ However, we must remember that the problem states $t \geq 0$ . Therefore, the negative value of $t$ is not valid for this problem.The only time when the particle is at rest for $t \geq 0$ is at $t = 3$ .

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