Read and draw conversion and travel graphs, interpret gradients as speed or acceleration, estimate instantaneous rates using tangents, and calculate distance from the area under a speed–time graph.
Subtopic 1
Use a straight line through the origin when the conversion factor is constant. Label both axes and units, choose a sensible scale, and read the required horizontal and vertical coordinates carefully.
A graph has 1 dollar equivalent to 0.8 euros. How many euros correspond to 65 dollars?
A conversion graph changes one unit or currency into another. If the conversion rate is constant, the graph is a straight line. To convert in one direction, start from one axis, move to the line and then across to the other axis.
A graph normally gives an estimate; calculation using the exact conversion rate is more accurate.
The exchange rate is £1 = $1.60. Give two points that can be used to draw the graph.
The graph passes through the origin: \((0,0)\).
£100 corresponds to $160, so another point is \((100,160)\).
Draw the straight line through these points.
Using the textbook graph, estimate the dollar value of £30.
From £30, move up to the line and across to the dollar axis.
The reading is about \(\boxed{\$48}\).
Using the same graph, estimate how many pounds are equivalent to $150.
Start at $150 on the vertical axis, move across to the line, then down to the pounds axis.
The graph gives approximately \(\boxed{£94}\).
Subtopic 2
On a distance–time graph, a horizontal part means stationary. A steeper rising line means faster travel; a falling line on a distance-from-start graph means returning towards the start.
A traveller moves from 0 to 30 km in 1 hour, waits 30 minutes and then goes from 30 to 50 km in 30 minutes. Find total average speed.
A travel graph plots time on the horizontal axis and distance from a fixed point on the vertical axis. The slope describes the motion.
On the textbook travel graph, what is happening from B to C?
The graph is horizontal, so the distance from A does not change.
The truck is \(\boxed{\text{stationary}}\).
Which stage is faster: A to B or C to D?
A to B: 10 km in 30 min.
C to D: 20 km in 30 min.
C to D has the steeper line and greater distance in the same time, so \(\boxed{\text{C to D is faster}}\).
A student runs 3 km in 15 minutes, waits for 5 minutes, then walks another 2 km in 20 minutes. State the three graph segments.
Subtopic 3
Speed is the gradient: vertical change divided by horizontal change. For a curved graph, draw a tangent at the required point, choose two widely spaced points on that tangent, and calculate its gradient.
A straight segment rises from 80 m at 10 s to 200 m at 25 s. Find speed.
The gradient of a distance–time graph gives the speed.
\[\text{speed}=\text{gradient}=\frac{\text{change in distance}}{\text{change in time}}\]
A straight-line graph rises 15 m in 20 s. Find the speed.
A traveller covers 10 km in 30 minutes. Find the speed in km/h.
30 minutes = 0.5 hours.
\[\text{speed}=\frac{10}{0.5}=\boxed{20\text{ km/h}}\]A tangent to a curved distance–time graph rises 30 m over a run of 15 s. Estimate the speed at the point of contact.
Make the tangent triangle as large as possible while keeping it on the tangent line. A larger triangle usually gives a more reliable gradient estimate.
Subtopic 4
Acceleration is speed change divided by time. A negative gradient represents deceleration; keep units consistent and label acceleration in \( ext{m/s}^2\).
A vehicle increases speed from 5 m/s to 29 m/s in 8 seconds. Calculate acceleration.
The gradient of a speed–time graph gives the acceleration.
\[\text{acceleration}=\frac{\text{change in speed}}{\text{change in time}}\]
Speed increases by 20 m/s in 20 s. Find the acceleration.
A car slows from 35 m/s to 0 in 25 s. Find its acceleration.
The deceleration is \(1.4\text{ m/s}^2\).
A car is travelling at 35 m/s and then decelerates at 1.4 m/s². What is its speed 10 s later?
Speed decrease:
\[1.4\times10=14\text{ m/s}\]\[35-14=\boxed{21\text{ m/s}}\]Subtopic 5
Distance equals area under a speed–time graph when speed is measured against time. Split the area into rectangles, triangles and trapezia; do not confuse this with a distance–time gradient.
A car accelerates from rest to 20 m/s in 10 seconds then continues at 20 m/s for 15 seconds. Find distance.
The area between a speed–time graph and the time axis gives the distance travelled.
An object travels at 30 m/s for 20 s. Find the distance.
Speed increases uniformly from 0 to 18 m/s over 20 s. Find the distance travelled.
The area is a triangle.
\[d=\frac12\times20\times18=\boxed{180\text{ m}}\]A car accelerates from 0 to 35 m/s in 20 s, stays at 35 m/s for 30 s, then slows to rest in 25 s. Find the total distance.
Attempt the questions before opening the answers. The later questions mix graph interpretation, gradients, acceleration and areas.
If $1 = €0.70, give two points suitable for drawing a conversion graph up to $100.
Use \((0,0)\) and \((100,70)\).
A conversion graph shows £20 ≈ NZ$65. Estimate NZ$ for £80 assuming a straight-line conversion.
£80 is 4 times £20, so NZ$ is approximately \(4\times65=\boxed{260}\).
What does a horizontal section of a distance–time graph mean?
The distance is not changing, so the object is \(\boxed{\text{stationary}}\).
A traveller covers 12 km in 30 min, then 12 km in 20 min. Which part has the steeper travel-graph line?
The same distance is covered in less time during the second stage, so the second line is steeper.
A straight distance–time graph rises from 4 m to 28 m while time increases from 2 s to 10 s. Find the speed.
A cyclist travels 15 km in 45 minutes. Find the speed in km/h.
45 min = 0.75 h.
\[\frac{15}{0.75}=\boxed{20\text{ km/h}}\]A tangent to a distance–time curve rises 24 m over 12 s. Estimate the instantaneous speed.
A distance–time curve becomes progressively steeper. What is happening to the speed?
The gradient is increasing, so the object is \(\boxed{\text{speeding up}}\).
Speed increases from 4 m/s to 16 m/s in 6 s. Find the acceleration.
A train slows from 18 m/s to 6 m/s in 8 s. Find the acceleration.
Deceleration = \(1.5\text{ m/s}^2\).
An object starts at 5 m/s and accelerates at 1.2 m/s² for 10 s. Find the final speed.
A car slows from 16 m/s with constant deceleration 0.8 m/s². How long does it take to stop?
An object moves at 12 m/s for 25 s. Find the distance travelled.
Speed rises uniformly from 0 to 20 m/s in 15 s. Find the distance.
A car’s speed increases uniformly from 10 m/s to 20 m/s over 12 s. Find the distance travelled.
Use trapezium area:
\[\frac12(10+20)(12)=\boxed{180\text{ m}}\]A car accelerates from rest to 24 m/s in 8 s, travels at 24 m/s for 15 s, then slows uniformly to rest in 12 s. Find the total distance.
A train accelerates uniformly from 0 to \(u\) m/s in 20 s, travels at \(u\) for 60 s, then slows uniformly to rest in 30 s. The total distance is 1100 m. Find \(u\).
A speed–time graph rises linearly from 6 m/s at \(t=0\) to 18 m/s at \(t=8\) s. Find (a) the acceleration and (b) the distance travelled in the 8 s.
a)
\[a=\frac{18-6}{8}=\boxed{1.5\text{ m/s}^2}\]b) Trapezium area:
\[d=\frac12(6+18)(8)=\boxed{96\text{ m}}\]I read the labels, units and value of each small square before calculating.
I can recognise moving, stationary and returning stages, and compare speeds from steepness.
I know gradient gives speed and I can use a tangent on a curve.
I know gradient gives acceleration and a negative gradient represents deceleration.
I split the area into rectangles, triangles or trapezia and add them accurately.
I convert minutes to hours or other units before calculating when required.