The flat part of the graph is the highest the line reaches.

Read across to the velocity axis: \(16\) m/s.

Acceleration is the gradient of a velocity-time graph.

In the first \(4\) seconds the velocity rises from \(0\) to \(16\) m/s:

\(\text{acceleration} = \frac{ 16 }{ 4 } = 4\)

The units are metres per second, per second, so the answer is \(4\ \text{m/s}^2\).

Over the last \(8\) seconds the velocity falls from \(16\) m/s to \(0\).

\(\text{deceleration} = \frac{ 16 }{ 8 } = 2\)

So the car decelerates at \(2\ \text{m/s}^2\). (Its acceleration is \(-2\ \text{m/s}^2\), because the velocity is decreasing.)

The speed stays at \(550\) mph for the whole flight, so the graph is a straight horizontal line at a height of \(550\).

It starts at time \(0\). The flight lasts 3 hours 45 minutes, which is \(3.75\) hours on this axis.

Draw the line from \((0, 550)\) to \((3.75, 550)\).

On a velocity-time graph the distance is the area under the line.

The steady part is a rectangle \(10\) s wide and \(16\) m/s high:

\(10 \times 16 = 160\) m

Split the area under the graph into a triangle, a rectangle and a triangle.

Speeding up: \(\frac{1}{2} \times 2 \times 16 = 16\) m

Steady velocity: \(10 \times 16 = 160\) m

Slowing down: \(\frac{1}{2} \times 8 \times 16 = 64\) m

\(16 + 160 + 64 = 240\) m

Deceleration is how steeply the line falls, which is the drop in velocity divided by the time it takes.

The velocity falls from \(12\) m/s to \(10\) m/s, a drop of \(2\) m/s, and it takes \(2\) seconds.

\(\frac{ 2 }{ 2 } = 1\)

So the van decelerates at \(1\ \text{m/s}^2\).

Distance is the area under the line.

Between \(4\) s and \(6\) s the shape is a trapezium. Its two parallel sides are the velocities \(12\) m/s and \(10\) m/s, and the distance between them is the \(2\) seconds.

\(\frac{1}{2} \times (12 + 10) \times 2 = 22\)

So the van travels \(22\) m.

The distance covered in the first 13 seconds is the area under the graph.

She speeds up for 2 s to 6 m/s (a triangle), then runs at 6 m/s for the next \(13 - 2 = 11\) s (a rectangle):

\(\frac{1}{2} \times 2 \times 6 + 11 \times 6 = 6 + 66 = 72\) m

The distance left is \(168 - 72 = 96\) m, at a steady 6 m/s:

\(96 \div 6 = 16\) s

Her finishing time is \(13 + 16 = 29\) s.

Ruth took 28.5 s, so the answer is No: Ella's 29 s compares with Ruth's 28.5 s.

The strips are \(1\) s wide, so the heights are read at \(t = 1\), \(2\), \(3\), \(4\) and the last at \(t = 5\). The heights read from the graph are \(2,\ 6,\ 12,\ 20,\ 30\) m/s.

Each trapezium has area \(\frac{1}{2} \times 1 \times (h_1 + h_2)\), giving \(4,\ 9,\ 16,\ 25\).

Total estimate: \(54\) m.

Distance is speed multiplied by time, so the area under a speed-time graph is the distance travelled.

The area represents the distance, in metres, that the object travels between \(t = 1\) s and \(t = 5\) s.

Acceleration is the gradient of the line over that stage, not over the whole journey — the graph has a different steepness before and after \(4\) seconds.

Between \(4\) s and \(6\) s the velocity rises from \(20\) m/s to \(26\) m/s, which takes \(2\) seconds.

\(\frac{ 26 - 20 }{ 2 } = 3\)

So the acceleration is \(3\ \text{m/s}^2\).

The distance is the area under the graph, and it splits into three pieces.

From \(0\) to \(4\) seconds it is a triangle, \(4\) s wide and \(20\) m/s high:

\(\frac{1}{2} \times 4 \times 20 = 40\) m

From \(4\) to \(6\) seconds it is a trapezium with parallel sides \(20\) m/s and \(26\) m/s, \(2\) s apart:

\(\frac{1}{2} \times (20 + 26) \times 2 = 46\) m

From \(6\) to \(12\) seconds the velocity is steady, so it is a rectangle:

\(6 \times 26 = 156\) m

\(40 + 46 + 156 = 242\) m

Draw the straight line from \((0, 0)\) to \((7, 8)\). The area under that line followed by the flat section splits into a triangle and a rectangle.

Triangle: \(\frac{1}{2} \times 7 \times 8 = 28\) m

Rectangle: \((11 - 7) \times 8 = 4 \times 8 = 32\) m

Together these make \(28 + 32 = 60\) m.

The curve is above the straight line, so the area under the curve is more than this.

So the distance is more than \(60\) metres.

Average acceleration is change in speed divided by time.

The speed goes from \(0\) m/s to \(8\) m/s and stays at \(8\) m/s after \(7\) seconds, so over the first \(10\) seconds the change in speed is \(8\) m/s.

\(\frac{ 8 }{ 10 } = 0.8\)

The units are metres per second per second, \(\text{m/s}^2\).

The distance in the first \(6\) seconds is the area under the drawn line, a trapezium with parallel sides \(16\) m/s and \(12\) m/s:

\(\frac{1}{2} \times (16 + 12) \times 6 = 84\) m

The distance left for the last \(6\) seconds is

\(150 - 84 = 66\) m

In the last \(6\) seconds the area under the graph is also a trapezium, with parallel sides \(12\) m/s and \(v\) m/s:

\(\frac{1}{2} \times (12 + v) \times 6 = 66\)

Multiply both sides by 2 and divide by \(6\):

\(12 + v = 22\)

\(v = 22 - 12 = 10\)

So Hafsa crosses the line at \(10\) m/s.

Sketching the velocity-time graph makes the three stages easy to see: a slope up, a flat section, and a slope down. The distance is the area underneath.

Speeding up is a triangle \(8\) s wide and \(12\) m/s high:

\(\frac{1}{2} \times 8 \times 12 = 48\) m

The steady stage is a rectangle:

\(90 \times 12 = 1080\) m

Slowing down is a triangle \(10\) s wide:

\(\frac{1}{2} \times 10 \times 12 = 60\) m

\(48 + 1080 + 60 = 1188\) m

Average speed is the total distance divided by the total time, not the average of the velocities in the question.

The whole journey takes \(8 + 90 + 10 = 108\) seconds.

\(\frac{ 1188 }{ 108 } = 11\) m/s

It is less than \(12\) m/s, as it has to be — the train was only travelling at \(12\) m/s for part of the time.