We can use a speed graph to calculate the distance travelled in a given time interval by finding the area under the line between the start and end times.

a)

The shape made by the speed curve, the line $x=0$, and the lines $t=4$ and $t=6$ seconds is a rectangle, so we can work out the area of this section by multiplying the width by the height.

The rectangle is $2$ seconds wide, and $\var{c1}$ ms^-1 high.

\begin{align}
\text{Area} &= \text{width} \times \text{height}\\
&= 2 \times\var{c1}\\
&=\simplify{2{c1}}\text{.}
\end{align}

So the distance covered in this two second interval is $\simplify{2{c1}}$ m.

b)

The shape made by the line and $x=0$ between $0$ and $2$ seconds forms a right-angled triangle with width $2$ and height $\var{b1}$.

\begin{align}
\text{Area}&=  \frac{1}{2}\times \text{width} \times \text{height}\\
&= \frac{1}{2} \times 2 \times \var{b1}\\
&=\var{b1} \text{.}
\end{align}

So therefore,the distance covered in this two second interval, and our answer, is $\simplify{{b1}}$ meters.

c)

The shape made by the speed curve and $x=0$ between $2$ and $4$ seconds forms a trapezium. This can be broken down in to a right angle triangle (let's call this $A$) and a rectangle (we'll call this $B$).

Triangle $A$ has width $2$ m and height $\var{c1}-\var{b1}$ ms^-1.

\begin{align}
A &=  \frac{1}{2}\times \text{width} \times \text{height}\\
&= \frac{1}{2}\times2 \times\ (\var{c1}-\var{b1})\\
&= \var{c1}-\var{b1}\\
&=\simplify{{c1}-{b1}}\text{.}
\end{align}

We can work out the area of the rectangle $B$ by multiplying its width, $2$ seconds, by its height, $\var{b1}$ ms^-1:

\begin{align}
B &=  \text{width} \times \text{height}\\
&= 2 \times(\var{c1}-\var{b1})\\
&=2 \times \simplify{{c1}-{b1}}\\
&=\simplify{2{c1-b1}}\text{.}
\end{align}

We can now work out the whole area under the line by adding these two areas together:

\begin{align}
\text{Area} &= A + B \\
&=\simplify{{c1}-{b1}} + \simplify{2{c1-b1}} \\
&=\simplify{2{c1-b1}+{c1}-{b1}} \text{.}
\end{align}

The distance covered in this interval is $\var{2(c1-b1)+c1-b1}$ m.

d)

Speed is the distance travelled per unit of time.

\begin{align}
\text{speed} &= \frac{\text{distance}}{\text{time}} \\[0.5em]
&= \frac{\var{area}}{10} \\[0.5em]
&=\simplify[!fractionNumbers]{{area/10}} \text{ ms}^{-1}\text{.}
\end{align}

a)

Imani's training run is $\var{km}$ km long. 

To convert $\var{km}$ km into metres, we multiply $\var{km}$ by $1000$. 

\[\var{km}\times1000= \var{km*1000}\text{ metres.}\]

b)

To convert $\var{km}$ km into miles, we multiply $\var{km}$ by the conversion rate given: $0.62$.

\[\begin{align}
\var{km}\times\frac{5}{8}&= \var{miles}\\
&=\var{dpformat(miles,0)}\text{ miles, rounded to the nearest integer.}
\end{align}\]

c)

The London Marathon is $26$ miles long. To convert to km we multiply by the inverse of the conversion rate given in part b):

\[ 26 \times \frac{1}{0.62} = 42\text{ miles, rounded to the nearest integer.} \]

Craig is $5$ft $\var{inches}$ inches tall.

a)

To find Craig's height into cm, we first convert it into inches. 

We can convert feet into inches by multiplying by $12$. So $5$ feet is

\[ 5\times12=60 \text{ inches.}\]

Therefore $5$ft $\var{inches}$ inches is 

\[
60+\var{inches} = \var{60+inches}\text{ inches.}
\]

We can now convert this to cm by multiplying by $2.54$.

\[\var{60+inches}\times2.54=\var{cm}\text{ cm, rounded to the nearest integer.}\]

b)

To convert centimetres into metres, we divide by $100$:

\[\var{cm}\div100=\var{dpformat(cm/100,2)} \text{ metres.}\]

Therefore, Craig is $\var{dpformat(cm/100,2)}$ metres tall. 

a)

To find the average speed of the runner in meters per second (m/s), we divide the distance covered by the runner (in metres) by the time taken for the runner to run this distance (in seconds).

\[
\begin{align}
\text{Average speed} &= \displaystyle\frac{\var{distance}}{\var{seconds}}\\
&= \var{distance/seconds}\\
&= \var{dpformat(distance/seconds,2)}\; \text{m/s} \; (\text{rounded to $2$ decimal places}).
\end{align}
\]

b)

We can convert the average speed of the runner that we calculated in a) in metres per second to kilometres per hour using the following two equivalences:

\[1\text{m} = \displaystyle\frac{1}{1000}\text{km},\]

\[
1 \; \text{second} = \displaystyle\frac{1}{60} \; \text{minutes} = \displaystyle\frac{1}{3600} \; \text{hours}.
\]

We know from a) that the average speed of the runner in m/s was $\var{dpformat(distance/seconds,5)}$ m/s ($5$ d.p), so to convert this speed to km/h we first need to convert metres to kilometres,

\[\var{dpformat(distance/seconds,5)} \; \text{m/s} = \var{dpformat(distance/seconds/1000,5)} \text{km/s} \; (5 \; \text{d.p})\]

Then we convert seconds to hours,

\[1 \; \text{second} = \displaystyle\frac{1}{3600} \; \text{hours}.\]

Now we have

\[\var{dpformat(distance/seconds,5)} \; \text{m/s} = \var{sigformat(distance/seconds/1000,5)} \; \text{kilometres per $\displaystyle\frac{1}{3600}$ hours}.\]

We want a rate per one hour, so we multiply by $3600$ to obtain a measurement in km/h:

\[\begin{align}\var{sigformat(distance/seconds/1000,5)} \; \text{kilometres per $\displaystyle\frac{1}{3600}$ hours} &= \var{siground(distance/seconds/1000,5)*3600} \; \text{km}/\text{h}\\&=\var{dpformat(distance/seconds*3.6, 2)} \; \text{km}/\text{h} \; (\text{rounded to $2$ decimal places}).\end{align}\]

Note that throughout this calculation we have rounded all figures to $5$ decimal places for convenience; when doing calculations which involve long decimals, you should always input the full figure into your calculator to avoid getting an incorrect answer due to rounding.

We are told that the track is $\var{kilometres}$km long and that the speed of the rabbit is $\var{speed}$m/s (metres per second).

Most people remember the relationship between speed, distance and time with the formula

\[\text{Average speed} = \frac{\text{Distance travelled}}{\text{Total time taken}}.\]

We can rearrange the formula for average speed to give us the formula for the time taken: 

\[\text{Total time taken} = \displaystyle\frac{\text{Distance travelled}}{\text{Average speed}}.\]

Firstly, we must convert the units of the length of the race from kilometres to metres. We know that $1\text{km} = 1000\text{m}$, therefore

\begin{align}
\var{kilometres}\text{km} &= (\var{kilometres} \times 1000)\text{m}\\
&= \var{{kilometres}*1000}\text{m}.
\end{align}

Therefore, the time taken for the rabbit to finish one lap is

\[ \begin{align} \displaystyle\frac{\var{{kilometres}*1000}}{\var{speed}} &= \var{({kilometres}*1000)/{speed}} \; \text{seconds}\\ &= \var{dpformat(({kilometres}*1000)/{speed}, 0)} \; \text{seconds} \; (\text{to the nearest second}).  \end{align}\]

a)

The gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).

\[ m = \frac{y_2-y_1}{x_2-x_1}=\frac{\simplify[!collectNumbers]{{yb}-{ya}}}{\simplify[!collectNumbers]{{xb}-{xa}}}=\frac{\simplify{{yb}-{ya}}}{\simplify{{xb}-{xa}}}=\simplify[simplifyFractions,unitDenominator]{({yb-ya})/({xb-xa})}\text{.}\]

b)

Rearranging the equation $y=mx+c$ for $c$ and using point A:

\[ c = y_1-mx_1 = \var{ya}-\var{m}\times\var{xa}=\simplify{{ya-m*xa}}\text{.}\]

We then check this against point $B$:

\[ y_2 = mx_2 + c = \simplify[fractionNumbers]{{m}{xb}+{c}}=\simplify{{m}*{xb}+{c}}\text{.}\]

b)

We now substitute the values for $m$ and $c$ into the equation of a straight line, $y=mx+c$,

\[y=\simplify[!noLeadingMinus,unitFactor]{{m} x+ {c}}\text{.}\]

c)

The gradient represents the vertical change (height in cm) per unit of the horizontal axis (days): the change in height of the sunflower per day.

d)

Substituting $x=\var{d}$ into the straight line equation, the height $y$ after $\var{d}$ days is

\begin{align}
y&=\simplify{{m}}x+\var{c}\\
&=\simplify[]{{m}{d}}+\var{c}\\
&=\var{m*d+c}\text{cm.}
\end{align}

e)

Substituting $x=\var{1826}$ into the straight line equation, the height $y$ after $1826$ days is

\begin{align}
y&=\simplify{{m}}x+\var{c}\\
&=\simplify[]{{m}1826} + \var{c}\\
&=\var{m*1826+c}\text{cm.}
\end{align}

Note that this is $\var{(m*1826+c)/100}$ metres. In 2014, a sunflower of $9.17$ metres was entered into the Guinness World Records as tallest sunflower.

f)

Possible reasons that the prediction will not be accurate are:

• from everyday observations and basic biology that a sunflower cannot keep growing forever, but our straight line equation would predict otherwise.
• the set of observations are limited in their time frame. In particular another relationship could easily fit the 2 data points. 

Invalid reasons that the prediction will not be accurate are:

• There can only be one straight line between two points; only one equation describes this.
• Based on common sense, we know that sunflowers do grow over time.