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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)-15t^(2)+27 t. 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 $x(t)=t^{3}-15 t^{2}+27 t$ . 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}-15 t^{2}+27 t

. Determine all values of

t

when the particle is at rest.

\newline

Answer:

t=

Find Velocity Derivative: To find when the particle is at rest, we need to determine when its velocity is zero. The velocity is the derivative of the position function $x(t)$ . Let's find the derivative of $x(t) = t^3 - 15t^2 + 27t$ . $v(t) = \frac{dx}{dt} = \frac{d}{dt} (t^3 - 15t^2 + 27t) = 3t^2 - 30t + 27$ .
Solve for Zero Velocity: Now we need to solve the equation $v(t) = 0$ to find the values of $t$ when the particle is at rest. $\newline$ $0 = 3t^2 - 30t + 27$ .
Simplify Equation: We can simplify the equation by dividing all terms by $3$ , which will not change the solutions. $0 = t^2 - 10t + 9$ .
Factor Quadratic Equation: Next, we factor the quadratic equation to find the values of $t$ . $(t - 9)(t - 1) = 0$ .
Find Values of $t$ : Setting each factor equal to zero gives us the values of $t$ when the particle is at rest. $t - 9 = 0$ or $t - 1 = 0$ . So, $t = 9$ or $t = 1$ .

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