Question

The position of a particle is given by the function x = 12t 3 - 9t 2 + 122 m, where t is in s. a. At what time or times is vx = 0 m/s? b. What are the particle’s position and its acceleration at this time(s)?

Short answer

(a) The times at which $v_x=0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$are $t=0s$and $t=3s$.

(b) The position and acceleration of the particle at $t=0s$are $12m$and $-18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$respectively.

The position and acceleration of the particle at $t=3s$are $-15m$and $18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$respectively.

Step-by-step solution

Given information

The position of the particle is given by the function $x=\left(2t^3-9t^2+12\right)m$.

Part (a)Time at which vx=0 ms.

The velocity is given by $v_x=\frac{dx}{dt}$.

As $v_x=0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$it implies that $\frac{dx}{dt}=0$.

Differentiate the given function with respect to $t$.

$$\begin{gathered} \frac{d\left(2t^3-9t^2+12\right)}{dt}=0 \\ \frac{d\left(2t^3\right)}{dt}-\frac{d\left(9t^2\right)}{dt}+\frac{d\left(12\right)}{dt}=0 \\ 2\left(3t^2\right)-9\left(2t\right)=0 \\ 6t^2-18t=0 \\ 6t\left(t-3\right)=0 \\ t=0ort=3 \end{gathered}$$

Therefore, the times at which$v_x=0$are$0s$and$3s$.

Part (b)Position of particle at t=0s.

To find the position of the particle at $t=0s$, substitute 0 for $t$into the given function.

$$\begin{gathered} x\left(att=0s\right)=2{\left(0\right)}^3-9{\left(0\right)}^2+12 \\ =12m \end{gathered}$$

Therefore, the position of the particle at$t=0s$is$12m$.

Acceleration of the particle at t=0s.

To find the acceleration of the particle, differentiate the given function twice.

$$\begin{gathered} a_x=\frac{d^{2}x}{d{t}^2} \\ =\frac{d^2\left(2t^3-9t^2+12\right)}{d{t}^2} \\ =12t-18 \end{gathered}$$

Substitute 0 for $t$into the obtained expression.

$$\begin{gathered} a_x\left(att=0s\right)=12\left(0\right)-18 \\ =-18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of the particle at$t=0s$is$-18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

The position of the particle at t=3s.

To find the position of the particle at $t=3s$, substitute 3 for $t$into the given function.

$$\begin{gathered} x=2{\left(3\right)}^3-9{\left(3\right)}^2+12 \\ =2\left(27\right)-9\left(9\right)+12 \\ =-15m \end{gathered}$$

Therefore, the position of the particle at $t=3s$is $-15m$.

The acceleration of the particle at t=3s.

To find the acceleration of the particle, differentiate the given function twice.

$$\begin{gathered} a_x=\frac{d^2\left(2t^3-9t^2+12\right)}{d{t}^2} \\ =12t-18 \end{gathered}$$

Substitute 3 for $t$into the obtained expression.

$$\begin{gathered} a_x\left(att=3s\right)=12\left(3\right)-18 \\ =18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of the particle at $t=3s$is $18\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.