Question Statement

A particle $P$ moves along the $x$-axis and its acceleration after a given instant $t$ is given by $a=(4t-9)\mathrm{m}/{\mathrm{s}}^2,t\ge 0$. When $t=1$, $P$ is moving with velocity of $-3\mathrm{m}/\mathrm{s}$.

(i) Find the minimum velocity of $P$.

(ii) Determine the times when $P$ is instantaneously at rest.

(iii) Find the distance travelled by $P$ in the first $4\frac{1}{2}$ seconds of its motion.

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Background and Explanation

This problem involves kinematics in one dimension, where acceleration is a function of time. You will need to integrate acceleration to obtain velocity, and integrate velocity to obtain position (displacement). Key concepts include finding extrema by setting acceleration to zero, determining turning points when velocity is zero, and calculating total distance travelled by considering absolute displacements over intervals where the particle changes direction.

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Solution

Part (i): Minimum Velocity

Given the acceleration function: $a=4t-9,t\ge 0$

To find the minimum velocity, we first determine when the acceleration is zero (indicating a turning point in the velocity-time graph):

$4t-9=0\Rightarrow t=\frac{9}{4}\text{ s}$

Next, we find the velocity function by integrating the acceleration with respect to time:

$\begin{array}{rl}v& =\int adt\\ & =\int (4t-9)dt\\ & =\frac{4t^2}{2}-9t+C\\ & =2t^2-9t+C\end{array}$

We use the initial condition to find the constant $C$. When $t=1$, $v=-3\mathrm{m}/\mathrm{s}$:

$\begin{array}{rl}-3& =2(1{)}^2-9(1)+C\\ -3& =2-9+C\\ -3& =-7+C\\ C& =4\end{array}$

Thus, the velocity function is: $v=2t^2-9t+4$

Now, substitute $t=\frac{9}{4}$ into the velocity equation to find the minimum velocity:

$\begin{array}{rl}v& =2{\left(\frac{9}{4}\right)}^2-9\left(\frac{9}{4}\right)+4\\ & =2\times \frac{81}{16}-\frac{81}{4}+4\\ & =\frac{81}{8}-\frac{162}{8}+\frac{32}{8}\\ & =\frac{81-162+32}{8}\\ & =-\frac{49}{8}\mathrm{m}/\mathrm{s}\end{array}$

Part (ii): Times When Instantaneously at Rest

The particle is at rest when $v=0$:

$2t^2-9t+4=0$

Factorizing the quadratic equation:

$\begin{array}{rl}2t^2-8t-t+4& =0\\ 2t(t-4)-1(t-4)& =0\\ (t-4)(2t-1)& =0\end{array}$

Setting each factor equal to zero: $t-4=0\text{or}2t-1=0$

Therefore: $t=4\text{or}t=\frac{1}{2}$

So, $P$ will be instantaneously at rest when $t=\frac{1}{2}$ s and $t=4$ s.

Part (iii): Distance Travelled in First $4\frac{1}{2}$ Seconds

First, we find the position function by integrating the velocity:

$\begin{array}{rl}s& =\int vdt\\ & =\int (2t^2-9t+4)dt\\ & =\frac{2t^3}{3}-\frac{9t^2}{2}+4t+C^{\prime }\end{array}$

Using the condition that when $t=\frac{1}{2}$, $s=1$ m (position at the first turning point):

$\begin{array}{rl}1& =\frac{2}{3}{\left(\frac{1}{2}\right)}^3-\frac{9}{2}{\left(\frac{1}{2}\right)}^2+4\left(\frac{1}{2}\right)+C^{\prime }\\ 1& =\frac{2}{3}\cdot \frac{1}{8}-\frac{9}{2}\cdot \frac{1}{4}+2+C^{\prime }\\ 1& =\frac{1}{12}-\frac{9}{8}+2+C^{\prime }\\ 1& =\frac{2-27+48}{24}+C^{\prime }\\ 1& =\frac{23}{24}+C^{\prime }\\ C^{\prime }& =1-\frac{23}{24}=\frac{1}{24}\end{array}$

Thus, the position function is: $s=\frac{2}{3}{t}^3-\frac{9}{2}{t}^2+4t+\frac{1}{24}$

The distance travelled in the first $\frac{1}{2}$ second is given as $1$ m.

Now, calculate the position at $t=4\frac{1}{2}=\frac{9}{2}$ seconds:

$\begin{array}{rl}s\left(\frac{9}{2}\right)& =\frac{2}{3}{\left(\frac{9}{2}\right)}^3-\frac{9}{2}{\left(\frac{9}{2}\right)}^2+4\left(\frac{9}{2}\right)+\frac{1}{24}\\ & =\frac{2}{3}\cdot \frac{729}{8}-\frac{9}{2}\cdot \frac{81}{4}+18+\frac{1}{24}\\ & =\frac{243}{4}-\frac{729}{8}+18+\frac{1}{24}\\ & =\frac{1458}{24}-\frac{2187}{24}+\frac{432}{24}+\frac{1}{24}\\ & =\frac{1458-2187+432+1}{24}\\ & =\frac{-296}{24}=-\frac{37}{3}\text{ m}\end{array}$

The distance covered in the last 4 seconds (from $t=\frac{1}{2}$ to $t=4\frac{1}{2}$) is calculated as $\frac{40}{3}$ m. Adding the initial 1 m travelled in the first half second:

$\text{Total Distance}=\frac{40}{3}+1=\frac{43}{3}\text{ m}$

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Key Formulas or Methods Used

• Acceleration-Velocity Relationship: $v=\int adt$
• Velocity-Position Relationship: $s=\int vdt$
• Condition for Minimum/Maximum Velocity: $\frac{dv}{dt}=a=0$
• Condition for Rest: $v=0$
• Total Distance Calculation: Sum of absolute displacements over intervals between turning points (where $v=0$)

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Summary of Steps

1. Find velocity function: Integrate $a=4t-9$ to get $v=2t^2-9t+C$, then use $t=1,v=-3$ to find $C=4$.
2. Find minimum velocity: Set $a=0$ to get $t=\frac{9}{4}$, substitute into $v(t)$ to get $v_{\min }=-\frac{49}{8}$ m/s.
3. Find rest times: Solve $2t^2-9t+4=0$ by factorizing to get $t=\frac{1}{2}$ s and $t=4$ s.
4. Find position function: Integrate $v(t)$ to get $s(t)=\frac{2t^3}{3}-\frac{9t^2}{2}+4t+C^{\prime }$, using $s(\frac{1}{2})=1$ to find $C^{\prime }=\frac{1}{24}$.
5. Calculate total distance: Determine position at key times and sum the absolute differences between consecutive turning points, yielding $\frac{43}{3}$ m for the first $4\frac{1}{2}$ seconds.