Distance travelled is equal to the area under the graph.

Find the total distance travelled by splitting the area under the graph into two or more shapes whose area can be calculated.

cie-igcse-20120-oct-nov-p2-tz3-q19b-working

Area under graph = area of trapezium + area of triangle

Area of trapezium $\begin{array}{rcl} \begin{array}{cl} = & \frac{v + (v + 10)}{2} \times 24 \end{array} & = & 12 (2 v + 10) \end{array}$

Area of triangle $\begin{array}{rcl} \begin{array}{cl} = & \frac{1}{2} \times (v + 10) \times (40 - 24) \end{array} \begin{array}{cl} = & 8 (v + 10) \end{array} \end{array}$

Area under the graph $\begin{array}{cl} = & 12 (2 v + 10) + 8 (v + 10) \end{array}$

Correct expression for one area [1]
Correct expression for both areas [1]

Note that the units on the graph are in metres, so you will need to convert 1.24 km to metres first.

$1.24 km = 1.24 \times 100 m = 1240 m$

Form an equation using the distance of 1240 m.

$12 (2 v + 10) + 8 (v + 10) = 1240$

Correct equation [1]

Simplify.

$\begin{array}{rcl} 24 v + 120 + 8 v + 80 & = & 1240 \\ 32 v + 200 & = & 1240 \end{array}$

Solve by subtracting 200 and dividing by 32.

$\begin{array}{rcl} 32 v + 200 & = & 1240 \\ 32 v & = & 1240 - 200 \\ 32 v & = & 1040 \\ v & = & \frac{1040}{32} \\ v & = & 32.5 \end{array}$

$v = 32.5$ m/s [1]

Real-Life Graphs (Cambridge O Level Maths)

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