Travel Graphs Worked ExamplesIGCSE Maths 0580

The answer must be in $\text{m/s}^{2}$, so convert the speed to $\text{m/s}$ first. Multiply by $1000$ for metres, divide by $3600$ for seconds.

Evaluate the conversion.

(a) Acceleration is the gradient of the speed-time graph: change in speed over time taken. Starting from rest means the initial speed is $0$.

(b) Distance is the area under the speed-time graph. From rest to $30\text{ m/s}$ in a straight line gives a triangle.

Substitute base $= 12\text{ s}$ and height $= 30\text{ m/s}$.

Sketch the shape: it is a trapezium. You can either split it into three pieces or use the trapezium formula. Splitting it is shown first.

The constant-speed section is a rectangle.

The deceleration section is another triangle.

(a) Add the three areas.

Check with the trapezium formula: the parallel sides are the constant-speed time ($30$) and the total time ($8 + 30 + 12 = 50$), and the height is the speed ($20$).

(b) Average speed is total distance divided by total time — not the average of the speeds.

First find the cyclist's head start. From $09:00$ to $09:40$ is $40$ minutes, which is $\frac{40}{60} = \frac{2}{3}$ of an hour.

From $09:40$ onwards, the car closes the gap at the relative speed — the difference between the two speeds.

Time to close a $10\text{ km}$ gap at $45\text{ km/h}$.

(a) The distance from the town is how far the car has gone in that time.

Check with the cyclist: total cycling time is $\frac{2}{3} + \frac{2}{9} = \frac{8}{9}$ hours.

(b) Convert $\frac{2}{9}$ of an hour into minutes and seconds.

Add that to the car's start time of $09:40$.

You cannot average the two speeds. Work out the time for each leg separately, using time $=$ distance $\div$ speed.

Now the return leg.

Add for the total time.

The total distance is there and back.

Apply the definition of average speed.

(a) Find the two areas you can calculate. The acceleration phase is a triangle.

The deceleration phase is also a triangle.

The middle section is a rectangle of unknown width $T$.

Set the sum of the three areas equal to the given total distance.

Simplify and solve.

(b) The total time is all three phases added together.

Average speed is total distance over total time.

The distance is the area under the speed-time graph. With four strips of width $h = 5$, use the trapezium rule.

Identify the parts: first $= 0$, last $= 24$, and the middle readings are $9$, $16$ and $21$.

Add the middle values inside the bracket first.

Substitute and simplify inside the square bracket.

Evaluate.

The speed is increasing but the increases are getting smaller ($+9, +7, +5, +3$), so the curve bends downwards. Each straight chord therefore lies below the curve.

(a) Speed is the gradient of a distance-time graph: distance travelled divided by time taken.

(b) The distance stays at $6\text{ km}$, so the graph is horizontal. A zero gradient means zero speed.

(c) The return leg runs from $11{:}30$ to $13{:}00$, a time of $1.5$ hours, covering $6\text{ km}$.

(a) Acceleration is the gradient. The cyclist starts at $4\text{ m/s}$, not from rest, so the change in speed is $10 - 4$.

(b) Distance is the area under the graph from $0$ to $30$ seconds. Split it into a trapezium ($0$ to $12\text{ s}$) and a rectangle ($12$ to $30\text{ s}$).

The rectangle is $18\text{ s}$ wide and $10\text{ m/s}$ high. Add the two areas.

(c) A horizontal line at $6\text{ m/s}$ crosses the graph twice: once while accelerating and once while slowing down. On the first section the speed rises by $0.5\text{ m/s}$ every second from $4\text{ m/s}$.

On the last section the speed falls from $10$ to $0$ in $8$ seconds, a deceleration of $\frac{10}{8} = 1.25\text{ m/s}^{2}$. Falling from $10$ to $6$ takes $4 \div 1.25$ seconds after $t = 30$.

(a) The gradient of a distance-time graph is the speed. The curve gets steeper during the first $8$ seconds, so the speed is increasing (the runner is accelerating).

(b) The speed at an instant is the gradient of the tangent at that point. Use two well-separated points on the tangent: $(2, 0)$ and $(6, 16)$.

(c) The graph is straight from $(8, 32)$ to $(12, 64)$, so the speed is constant and equal to the gradient.

(d) Average speed is total distance $\div$ total time. The runner is stationary from $12$ to $15$ seconds, but that time still counts.

(a) The speeds are equal where the graphs cross. $A$'s acceleration is $\frac{24}{20} = 1.2\text{ m/s}^{2}$, so its speed is $15$ when:

(b) Distance is the area under each graph. Up to $12.5$ seconds, $A$'s graph is below $B$'s the whole time, so the area under $A$ is smaller: $A$ has travelled less far and is still behind. (Equal speed is not the same as equal distance.)

(c) Compare the areas up to $t = 20$. Under $A$ it is a triangle; under $B$ a rectangle.

$240 < 300$, so $A$ is still $60\text{ m}$ behind after $20$ seconds.

(d) After $20$ seconds $A$ travels at a constant $24\text{ m/s}$. Write each car's distance at time $t$ (for $t > 20$) and set them equal.

Solve for $t$.

Use the exact value $t = \frac{80}{3}$ to find the distance, then check with car $A$: $240 + 24 \times \frac{20}{3} = 240 + 160 = 400$ ✓.

(a) The scooter slows from $9\text{ m/s}$ to rest between $t = 24$ and $t = 36$. Deceleration is the size of the gradient of that straight section.

(b) At $t = 4$ the speed is $5\text{ m/s}$. Rule a straight line that touches the curve at $(4, 5)$ without crossing it. A well-drawn tangent passes close to $(0, 1)$ and $(8, 9)$.

The rate of increase of speed is the gradient of the tangent. Use two points on the tangent that are far apart.

(As this is an estimate from a drawn tangent, any answer from about $0.9$ to $1.1$ would be reasonable.)

(c) From $t = 18$ to $t = 36$ is $18$ seconds. The distance is the area under the graph: a rectangle from $18$ to $24$ and a triangle from $24$ to $36$.

Mean speed $=$ total distance $\div$ total time.

(a) Acceleration is the gradient of the first section: the change in speed divided by the time taken.

Multiply both sides by $40$, then rearrange.

(b) Braking runs from $t = 100$ to $t = 130$, so it lasts $30$ seconds and the speed falls by $24\text{ m/s}$.

(c) Distance is the area under the graph. The graph does not start at zero, so the first section is a trapezium; then a rectangle ($40$ to $100$ s) and a triangle ($100$ to $130$ s).

Evaluate each area.

Convert metres to kilometres by dividing by $1000$. The answer is exact, so do not round it.