The graph reaches its first high point at \(\left(2,\ 8\right)\).

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The flat interval runs from time 2 to time 3.

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Eight kilometres are covered in two hours.

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Time 4 is halfway through the straight return segment.

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The outward and return sections each cover 8 km. Include the one-hour stop in the five-hour total.

16 km; 3.2 \(\mathrm{km/h}\).

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16 km; 3.2 \(\mathrm{km/h}\).

The meaning depends on the vertical-axis variable.

The first means constant non-zero speed. The second means no change in distance, so the object is stationary along the stated path.

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The first means constant non-zero speed. The second means no change in distance, so the object is stationary along the stated path.

The rising section gains \(3\text{ m/s}\) each second. The final section loses \(6\text{ m/s}\) each second from time 5.

\(\frac{4}{3}\) s and \(\frac{16}{3}\) s.

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\(\frac{4}{3}\) s and \(\frac{16}{3}\) s.

Outward time is 2 h and return time is 1 h. Divide 12 km by 3 h.

4 \(\mathrm{km/h}\). The two speeds apply for different durations.

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4 \(\mathrm{km/h}\). The two speeds apply for different durations.

The allowed total duration is \(\frac{10}{2.5}=4\) hours. Subtract the moving time.

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B has distance \(2\left(t-1\right)\). On A’s return segment, A has distance \(20-4t\). Equate them, then check \(3\le t\le 5\). No earlier segment gives a valid meeting time after \(t=1\). The elapsed time since B started is \(t-1=\frac{8}{3}\) hours.

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