Question: A bus left Town $A$ at 08 00 and travelled $80$ km to Town $B$, waited there, then returned. Its distance-time graph shows the return leg as a straight line from $(09\,10,\ 80\text{ km})$ down to $(10\,30,\ 0)$. Find the speed of the return trip, and explain what the horizontal part of the graph shows.

Step 1. On a distance-time graph, speed = gradient. Read the return leg's rise and run:

distance covered $= 80$ km; time taken $= 09\,10 \to 10\,30 = 1$ h $20$ min $= \dfrac{4}{3}$ h.

Step 2.

$$\text{Speed} = \dfrac{80}{\;\frac{4}{3}\;} = 60 \text{ km/h}$$

Step 3. The horizontal section (constant distance as time passes) means the bus is stationary — that is the wait at Town $B$.

Step 4. Reading the leg's direction: distance from $A$ decreasing means travelling back towards $A$; the graph sloping down does not mean slowing down — its steepness, not its direction, carries the speed.

⚠ Watch out: Convert the clock interval properly ($1$ h $20$ min is $\frac{4}{3}$ h, not $1.2$ h) — dividing by $1.2$ gives the plausible-but-wrong $66.7$. Horizontal = stationary on a distance-time graph; on a speed-time graph horizontal means constant speed. Confusing the two graphs is the classic error.

Question: A body's speed-time graph over $5$ minutes rises from rest to a maximum speed $V$, holds it, then falls to rest — the first $2.5$ minutes forming a trapezium with parallel sides $2.5$ min and $1.5$ min. Given the distance in the first $2.5$ minutes is $60$ m, find $V$; then find the deceleration in the last $2$ minutes.

[diagram in original sheet: trapezium-shaped speed-time graph, 0–5 min]

Step 1. On a speed-time graph, distance = area under the graph. The first $2.5$ min is a trapezium:

$$60 = \dfrac{1}{2}(2.5 + 1.5)\,V$$

Step 2. Solve:

$$60 = 2V \;\Rightarrow\; V = 30 \text{ m/min}$$

Step 3. Acceleration = gradient. The last $2$ minutes drop from $30$ to $0$:

$$\text{gradient} = \dfrac{0 - 30}{2} = -15$$

Step 4. State it as a deceleration (positive number, direction in the word):

$$\text{Deceleration} = 15 \text{ m/min}^2$$

⚠ Watch out: The two graph types swap meanings — speed-time: gradient = acceleration, area = distance; distance-time: gradient = speed, area = nothing. Check the axis label before computing anything. Deceleration of $15$, or acceleration of $-15$ — never "deceleration $= -15$".