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}

Frequency is the number of times that an event occurs (in this case, obtaining a certain score) within a particular experiment (in this case, the exam).

We are given enough data that keeping a mental count of the frequencies is difficult:

Therefore, a better way to approach this problem would be to add an extra column to the frequency table, as illustrated below, to tally the number of students with each score. The number of tally marks for each score can then be read off and insterted into the frequency column.

a)

Organising the data in a frequency table helps to make mistakes less likely when calculating statistics from our data, summarising the responses all in one place with fewer numbers.

Each row of the frequency column gives the number of students with the corresponding number of siblings.

Always remember to check whether your frequency column adds up to the total (here, it is $30$) to make sure you have not left out any responses.

b)

Mean

The mean number of siblings is the total number of siblings, $\sum x$, divided by the number of students in the sample, $n$.

\begin{align}  
\sum x &= 0 \times \var{freq[0]} + 1 \times \var{freq[1]} + 2 \times \var{freq[2]} + 3 \times \var{freq[3]} + 4 \times \var{freq[4]} + 5 \times \var{freq[5]} + 6 \times \var{freq[6]}
\\
&= 0 + \var{1*freq[1]} + \var{2*freq[2]} + \var{3*freq[3]} + \var{4*freq[4]} + \var{5*freq[5]} + \var{6*freq[6]} \\&= \var{sum(a)} \text{.}
\end{align}

The total number of students $n$ is $30$.

Therefore the mean is

\begin{align}
\bar{x} &= \frac{\sum x}{n} \\
&= \frac{\var{sum(a)}}{30} \\
&= \var{mean} \text{.}
\end{align}

Rounding the answer to 2 decimal places, we get $\var{precround(mean, 2)}$.

Mode

The mode is the value with the highest frequency. Here, the mode is $\var{mode}$ siblings, with frequency $\var{freq[mode]}$.

Median

The median is the "middle" value in the sample, when arranged in numerical order.

Since $n = 30$, we have two middle values in this data (15^th and 16^th place). We can count from the top of the table until we locate rows where these middle values lie, as the numbers in the table are already sorted by order.

Here, both $15$^th and $16$^th value lie in the row $\var{asa[14]}$.

As $15$^th value $= 16$^th value $= \var{asa[14]}$, the median is $\var{asa[14]}$.