Question Statement

A particle $P$ moves along the positive $x$-axis. At time $t$ seconds the velocity of $P$ is $v=(3t^2-4t+3)\text{ m/s}$. When $t=0$, $P$ is at the origin $O$. Find the distance of $P$ from $O$ when $P$ is moving with minimum velocity.

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Background and Explanation

This problem combines differential and integral calculus with kinematics. To find when velocity is minimized, we use the fact that at a minimum, the derivative (acceleration) equals zero. We then integrate the velocity function to obtain the position function and evaluate it at the critical time.

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Solution

Finding the Time of Minimum Velocity

The velocity function is given by:

$v=3t^2-4t+3$

Velocity is minimized when the acceleration equals zero (since the coefficient of $t^2$ is positive, the parabola opens upward, ensuring this critical point is indeed a minimum).

First, find the acceleration by differentiating velocity with respect to time:

$\begin{array}{rl}a& =\frac{dv}{dt}\\ & =\frac{d}{dt}(3t^2-4t+3)\\ & =3\times 2t-4\\ & =6t-4\end{array}$

Set acceleration equal to zero to find the critical time:

$\begin{array}{rl}6t-4& =0\\ 6t& =4\\ t& =\frac{4}{6}=\frac{2}{3}\text{ seconds}\end{array}$

Finding the Position Function

To find the distance from the origin, we integrate the velocity function to obtain the displacement $s$:

$\begin{array}{rl}s& =\int vdt\\ & =\int (3t^2-4t+3)dt\\ & =\frac{3t^3}{3}-\frac{4t^2}{2}+3t+C\\ & =t^3-2t^2+3t+C\end{array}$

Applying Initial Conditions

We are given that when $t=0$, the particle is at the origin, so $s=0$:

$\begin{array}{rl}0& =(0{)}^3-2(0{)}^2+3(0)+C\\ 0& =C\end{array}$

Therefore, the position function is:

$s=t^3-2t^2+3t$

Calculating the Distance at Minimum Velocity

Substitute $t=\frac{2}{3}$ into the position function:

$\begin{array}{rl}s& ={\left(\frac{2}{3}\right)}^3-2{\left(\frac{2}{3}\right)}^2+3\left(\frac{2}{3}\right)\\ & =\frac{8}{27}-2\left(\frac{4}{9}\right)+2\\ & =\frac{8}{27}-\frac{8}{9}+2\\ & =\frac{8}{27}-\frac{24}{27}+\frac{54}{27}\\ & =\frac{8-24+54}{27}\\ & =\frac{38}{27}\text{ m}\end{array}$

Since the particle started at the origin ($s=0$), the distance from $O$ is:

$\text{Distance}=\frac{38}{27}-0=\frac{38}{27}\text{ m}$

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Key Formulas or Methods Used

• Acceleration: $a=\frac{dv}{dt}$ (derivative of velocity)
• Condition for minimum velocity: $\frac{dv}{dt}=0$ (acceleration equals zero)
• Position from velocity: $s=\int vdt$ (integration)
• Power rule for integration: $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$

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Summary of Steps

1. Differentiate the velocity function $v=3t^2-4t+3$ to find acceleration $a=6t-4$
2. Set acceleration to zero: $6t-4=0$ and solve to get $t=\frac{2}{3}$ seconds
3. Integrate velocity to find position: $s=t^3-2t^2+3t+C$
4. Apply initial condition ($t=0,s=0$) to find $C=0$
5. Substitute $t=\frac{2}{3}$ into the position function and simplify to get $s=\frac{38}{27}$ m