Question Statement

An object moves in a straight line according to the position function given below. If $S$ is measured in centimetres, find the distance travelled by the object in the indicated time interval:

(i) $S(t)=t^2-2t$ ; $[0,5]$

(ii) $S(t)=t^3-3t^2-9t$ ; $[0,4]$

(iii) $S(t)=6\sin \pi t$ ; $[1,3]$

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Background and Explanation

Distance traveled is the total path length covered, which requires integrating the absolute value of velocity $|v(t)|$ over the time interval. Since velocity $v(t)=S^{\prime }(t)$ may change sign (indicating direction reversal), we must identify where $v(t)=0$ (turning points) and integrate separately over subintervals where velocity maintains a consistent sign.

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Solution

Part (i): $S(t)=t^2-2t$ on $[0,5]$

First, find the velocity function by differentiating position:

$V(t)=\frac{dS}{dt}=2t-2$

Find when the object changes direction by setting $V(t)=0$:

$2t-2=0⟹t=1$

Analyze the sign of velocity:

• For $t\in (0,1)$: $V(t)=2t-2<0$ (moving backward)
• For $t\in (1,5)$: $V(t)=2t-2>0$ (moving forward)

Since velocity changes sign at $t=1$, we calculate distance by integrating the absolute value:

$\text{Distance}={\int }_0^5|V(t)|dt=-{\int }_0^{1}V(t)dt+{\int }_1^{5}V(t)dt$

Substitute $V(t)=2t-2$:

$=-{\int }_0^1(2t-2)dt+{\int }_1^5(2t-2)dt$

Compute the integrals separately:

$=-{\left[t^2-2t\right]}_0^1+{\left[t^2-2t\right]}_1^5$

$=-\left[(1-2)-(0-0)\right]+\left[(25-10)-(1-2)\right]$

$=-(-1)+\left[15-(-1)\right]$

$=1+16=\boxed{17\text{ cm}}$

Part (ii): $S(t)=t^3-3t^2-9t$ on $[0,4]$

Find the velocity function:

$V(t)=\frac{dS}{dt}=3t^2-6t-9$

Find critical points where $V(t)=0$:

$3t^2-6t-9=0$ $3(t^2-2t-3)=0$ $(t-3)(t+1)=0$

This gives $t=3$ or $t=-1$. Since $t=-1\notin [0,4]$, we only consider $t=3$.

Analyze the sign of $V(t)=3(t-3)(t+1)$:

• For $t\in (0,3)$: $V(t)<0$ (negative)
• For $t\in (3,4)$: $V(t)>0$ (positive)

Calculate distance:

$\text{Distance}={\int }_0^4|V(t)|dt=-{\int }_0^3(3t^2-6t-9)dt+{\int }_3^4(3t^2-6t-9)dt$

Compute the antiderivative $\int (3t^2-6t-9)dt=t^3-3t^2-9t$:

$=-{\left[t^3-3t^2-9t\right]}_0^3+{\left[t^3-3t^2-9t\right]}_3^4$

Evaluate the first integral: $-\left[(27-27-27)-(0-0-0)\right]=-(-27)=27$

Evaluate the second integral: $\left[(64-48-36)-(27-27-27)\right]=(-20)-(-27)=7$

Total distance: $=27+7=\boxed{34\text{ cm}}$

Part (iii): $S(t)=6\sin \pi t$ on $[1,3]$

Find the velocity function using the chain rule:

$V(t)=\frac{dS}{dt}=6\cos (\pi t)\cdot \pi =6\pi \cos (\pi t)$

Find when velocity equals zero:

$6\pi \cos (\pi t)=0⟹\cos (\pi t)=0$

$\pi t=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\dots$

$t=\frac{1}{2},\frac{3}{2},\frac{5}{2},\dots$

Within the interval $[1,3]$, the relevant critical points are $t=\frac{3}{2}$ and $t=\frac{5}{2}$.

Analyze the sign of $V(t)=6\pi \cos (\pi t)$ in each subinterval:

• For $t\in \left(1,\frac{3}{2}\right)$: $\cos (\pi t)<0⟹V(t)<0$
• For $t\in \left(\frac{3}{2},\frac{5}{2}\right)$: $\cos (\pi t)>0⟹V(t)>0$
• For $t\in \left(\frac{5}{2},3\right)$: $\cos (\pi t)<0⟹V(t)<0$

Set up the distance integral:

$\text{Distance}={\int }_1^3|V(t)|dt=-{\int }_1^{3/2}V(t)dt+{\int }_{3/2}^{5/2}V(t)dt-{\int }_{5/2}^{3}V(t)dt$

Substitute $V(t)=6\pi \cos (\pi t)$:

$=-6\pi {\int }_1^{3/2}\cos (\pi t)dt+6\pi {\int }_{3/2}^{5/2}\cos (\pi t)dt-6\pi {\int }_{5/2}^3\cos (\pi t)dt$

Integrate $\int \cos (\pi t)dt=\frac{\sin (\pi t)}{\pi }$:

$=-6{\left[\sin (\pi t)\right]}_1^{3/2}+6{\left[\sin (\pi t)\right]}_{3/2}^{5/2}-6{\left[\sin (\pi t)\right]}_{5/2}^3$

Evaluate using $\sin (\pi )=0$, $\sin (3\pi /2)=-1$, $\sin (5\pi /2)=1$, $\sin (3\pi )=0$:

$=-6\left(-1-0\right)+6\left(1-(-1)\right)-6\left(0-1\right)$

$=-6(-1)+6(2)-6(-1)$

$=6+12+6=\boxed{24\text{ cm}}$

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Key Formulas or Methods Used

• Velocity from position: $v(t)=\frac{dS}{dt}$ (derivative of position function)
• Distance traveled formula: $\text{Distance}={\int }_a^b|v(t)|dt$
• Handling absolute value: Split integral at points where $v(t)=0$; negate integrals where $v(t)<0$
• Power rule for integration: $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$
• Chain rule for trigonometric functions: $\frac{d}{dt}\sin (\pi t)=\pi \cos (\pi t)$ and $\int \cos (\pi t)dt=\frac{\sin (\pi t)}{\pi }$

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Summary of Steps

1. Differentiate the position function $S(t)$ to obtain velocity $V(t)$
2. Find critical points by solving $V(t)=0$ within the given interval
3. Test intervals between critical points to determine where $V(t)$ is positive or negative
4. Set up the distance integral as $\int |V(t)|dt$, splitting into separate integrals at each critical point
5. Apply absolute value by multiplying negative-velocity intervals by $-1$
6. Evaluate each definite integral and sum the results to obtain total distance