a)

We can analyse the nature of each line to find out which line resembles speed and which resembles acceleration. 

The speed line should be distinguishable from the acceleration line based on the fact speed is always positive and has a continuous function (a line where the points are joined). This suggests the blue line with circular points represents speed.

Where as acceleration can often be discontinuous and can be negative where speeds are decreasing. This suggests that the green line with diamond shaped points is the acceleration line.

b)

Speed is a scalar quantity without direction so speed is positive no matter the direction it travels in. Therefore our answer is "no- a car can never have a negative speed.".

c)

By defintion, when an object is slowing down it has a negative acceleration as the acceleration measures that change in speed. Therefore, our answer is "yes- a car could have negative acceleration.".

d)

To calculate acceleration we use the point at $0$ seconds, with a $0$ m/s speed and the point at $2$ seconds with a $\var{b1}$ m/s speed and we find the gradient of the line between them. To find the gradient of a straight line we can use:

\[
\begin{align}
\text{Gradient } &= \frac{y_1-y_2}{x_1-x_2}\\
&= \frac{\var{b1-a1}}{2-0}\\
&=\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{b1-a1}/2} \text{.}
\end{align}
\]

We can then compare the graph to our this answer in order to check that our calculation is correct and we see that we the same answer as was given.

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}

We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

We can find upper and lower bounds for the given measurements. Any values we go on to calculate will also be uncertain and have upper and lower bounds.

To find bounds for a given measurement, we divide the degree of accuracy by 2 and subtract or add this to our estimate to get lower and upper bounds respectively.

For example, $52$ rounded to the nearest integer has a lower bound of $51.5$ and an upper bound of $52.5$.

a)

The distance travelled is given by

\[ d = \text{Average speed} \times \text{Time taken} \]

We find bounds for speed and time first.

Lower bound for speed: $\var{speed} - 0.5 = \var{speed - 0.5} \text{ km/h}$

Upper bound for speed: $\var{speed} + 0.5 = \var{speed + 0.5} \text{ km/h}$

First, note that the speed is given in km/h and we want to find the distance in km. We will convert the given time into hours. 

Lower bound for time taken:

\begin{align} \var{atime} - 0.5 &= \var{atime - 0.5} \text{ min} \\[0.5em]
&= \frac{\var{atime - 0.5}}{60} \text{ h}
\end{align}

Upper bound for time taken:

\begin{align} \var{atime} + 0.5 &= \var{atime + 0.5} \text{ min} \\[0.5em]
&= \frac{\var{atime + 0.5}}{60} \text{ h}
\end{align}

Since we're multiplying the speed and time together, the lower bound for distance is the slowest speed multiplied by the shortest time:

\begin{align}
\text{Lower bound} &= \text{lower bound for speed} \times \text{lower bound for time}\\
&= \var{speed - 0.5} \times \frac{\var{(atime - 0.5)}}{60} \\
&= \var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \text{ km} \quad \text{(rounded to 2 decimal places).}
\end{align}

The upper bound for distance is the fastest speed multiplied by the longest time:

\begin{align}
\text{Upper bound} &= \text{upper bound for speed} \times \text{upper bound for time} \\
&= \var{speed + 0.5} \times \frac{\var{atime + 0.5}}{60} \\
&= \var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \text{ km} \quad \text{(rounded to 2 decimal places).}
\end{align}

Hence,

\[\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \leq d \lt \var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \text{.}\]

b)

We're told the speed and the distance travelled, so the time taken is given by

\[ t = \frac{\text{Distance travelled}}{\text{Average speed}} \]

We found upper and lower bounds for Tia's average speed above.

The distance of the evening run is given to the nearest kilometre, so we can compute bounds as follows:

Lower bound for distance: $\var{distance} - 0.5 = \var{distance - 0.5} \mathrm{km}$

Upper bound for distance: $\var{distance} + 0.5 = \var{distance + 0.5} \mathrm{km}$

The upper bound for the time taken is the longest distance divided by the slowest speed:

\begin{align}
\text{Upper bound} &= \text{upper bound for distance} \div \text{lower bound for speed} \\
&= \var{distance + 0.5} \div \var{speed - 0.5} \\
&= \var{(distance + 0.5)/(speed - 0.5)} \text{ hours.}
\end{align}

We're asked for the answer in minutes, to two decimal places.

\begin{align} 
\var{(distance + 0.5)/(speed - 0.5)} \text{ hours} &= \var{(distance + 0.5)/(speed - 0.5)}\times 60  \text{ min} \\
&= \var{precround((distance + 0.5)/(speed - 0.5)*60, 2)} \text{ minutes} \quad \text{(rounded to 2 decimal places)}
\end{align}

The lower bound for time is the shortest distance divided by the fastest speed:

\begin{align}
\text{Lower bound} &= \text{lower bound for distance} \div \text{upper bound for speed} \\
&= \var{distance - 0.5} \div \var{speed + 0.5} \\
&= \var{(distance - 0.5)/(speed + 0.5)}  \text{ hours.}
\end{align}

Converting into minutes, to two decimal places:

\begin{align} 
\var{(distance - 0.5)/(speed + 0.5)} \text{ hours} &= \var{(distance - 0.5)/(speed + 0.5)}\times 60  \text{ min} \\
&= \var{precround((distance - 0.5)/(speed + 0.5)*60, 2)} \text{ min.}
\end{align}

Therefore, we cannot confidently say Tia's time was less than 61 minutes as the upper bound for her time, $\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)}$ minutes, is above this threshold.

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

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