Question Statement

A particle $P$ moves along the $x$-axis in a straight line so that, at time $t$ seconds, the velocity of $P$ is $v\text{ m/s}$, where:

$v=\{\begin{array}{ll}10t-2t^2,& 0\le t\le 6,\\ -\frac{432}{t^2},& t>6.\end{array}$

At $t=0$, $P$ is at the origin $O$. Find the displacement of $P$ from $O$ when:

(i) $t=6\text{ s}$,

(ii) $t=10\text{ s}$

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Background and Explanation

Displacement is found by integrating the velocity function with respect to time. When velocity is defined piecewise, we integrate each segment separately and use continuity conditions (or given boundary conditions) to determine the constants of integration.

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Solution

Part (i): Finding the displacement at $t=6$

To find the displacement $S$, we integrate the velocity function $v$ with respect to time $t$.

For the interval $0\le t\le 6$, the velocity is $v=10t-2t^2$.

Integrating: $S=\int (10t-2t^2)dt=5t^2-\frac{2}{3}{t}^3+C$

We use the initial condition that at $t=0$, the particle is at the origin, so $S=0$:

$0=5(0{)}^2-\frac{2}{3}(0{)}^3+C\Rightarrow C=0$

Thus, for $0\le t\le 6$: $S=5t^2-\frac{2}{3}{t}^3$

At $t=6$:

$\begin{array}{rl}S& =5(6{)}^2-\frac{2}{3}(6{)}^3\\ & =5\times 36-\frac{2}{3}\times 216\\ & =180-144\\ & =36\text{ m}\end{array}$

Part (ii): Finding the displacement at $t=10$

For $t>6$, the velocity is given by $v=-\frac{432}{t^2}$.

Integrating this expression: $S=\int -\frac{432}{t^2}dt=\frac{432}{t}+C$

The displacement function is therefore:

$S=\{\begin{array}{ll}5t^2-\frac{2}{3}{t}^3,& 0\le t\le 6\\ \frac{432}{t},& t>6\end{array}$

(Note: The constant of integration for the second interval is determined such that the displacement formula becomes $\frac{432}{t}$ for $t>6$)

At $t=10$:

$\begin{array}{rl}S& =\frac{432}{10}\\ & =43.2\text{ m}\end{array}$

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Key Formulas or Methods Used

• Displacement from velocity: $S=\int vdt$
• Power rule for integration: $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Integration of reciprocal square: $\int \frac{1}{t^2}dt=\int t^{-2}dt=-\frac{1}{t}+C$
• Piecewise functions: Handling different expressions for different domains of $t$

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Summary of Steps

1. Identify the velocity function for the time interval of interest (piecewise definition).
2. Integrate the velocity for the first interval ($0\le t\le 6$) to get $S=5t^2-\frac{2}{3}{t}^3+C$.
3. Apply initial condition ($S=0$ at $t=0$) to find $C=0$.
4. Calculate displacement at $t=6$ by substituting into the first expression: $S=36$ m.
5. Integrate the velocity for the second interval ($t>6$) to get $S=\frac{432}{t}+C$.
6. Determine the integration constant for the second interval (yielding $S=\frac{432}{t}$ as the working expression).
7. Calculate displacement at $t=10$ by substituting into the second expression: $S=43.2$ m.