01 · Rectangle
Velocity is −4 m/s for 5 seconds. Find displacement and distance.
Hint
A rectangle has magnitude 4 × 5.
Worked solution
Displacement = −20 m and distance = 20 m.
Understand · explore · practise
Displacement and distance from velocity–time areas
Use signed velocity–time areas for displacement and absolute areas for distance, splitting regions at reversals.
Before you startTriangle and trapezium areas, signed velocity and displacement.
01 / Area accumulates displacement
The sign comes from velocity, not from a negative length.
Area under a velocity–time graph gives signed displacement.
Add areas above the time axis and subtract the magnitudes below it. For total distance, add both magnitudes. The area has units (m/s) × s = m.
Signed area and total distanceExplore
Velocity v = 6 − 2t m/s, with right positive. Choose the end time. The shaded area above the axis adds displacement; the area below subtracts displacement.
02 / Use geometry when each stage is straight
Use actual durations as widths.
Velocity rises uniformly from 2 to 8 m/s over 3 seconds.Worked example
Displacement = ½(2 + 8) × 3 = 15 m
The region is a trapezium.
Distance = 15 m
Velocity stays positive, so there is no reversal.
02 · Triangle
Velocity rises uniformly from rest to 10 m/s in 6 seconds. Find displacement.
Hint
Use half × width × height.
Worked solution
Displacement = ½ × 6 × 10 = 30 m.
03 · Offset clocks
Between t = 4 s and t = 9 s, velocity rises linearly from 3 to 7 m/s. Find displacement.
Hint
The interval lasts 5 seconds.
Worked solution
Displacement = ½(3 + 7) × 5 = 25 m.
03 / Keep the sign of an entire negative region
Geometrical area is a magnitude; signed area records direction.
04 · Below the axis
Velocity changes uniformly from −2 to −8 m/s over 4 seconds. Find displacement and distance.
Hint
Use the magnitudes for a trapezium, then assign its sign.
Worked solution
The area magnitude is ½(2 + 8) × 4 = 20 m. Displacement = −20 m; distance = 20 m.
05 · Zero area
A particle has v = 0 for 7 seconds. What does the graph contribute?
Hint
The height is zero.
Worked solution
It contributes zero displacement and zero distance, but the 7 seconds still count in a whole-journey average.
04 / Split where velocity is zero
Do this before calculating total distance.
Watch: cancellation changes displacement, not distance
Pause, replay or seek freely. The notes explain the same idea and stay in view.
06 · Crossing time
For v = 6 − 2t, find when motion reverses.
Hint
Set v = 0.
Worked solution
6 − 2t = 0 gives t = 3 s. Velocity is positive before and negative afterwards.
07 · Full interval
For that model on 0 ≤ t ≤ 6, find displacement and distance.
Hint
There are two triangles of base 3 and height 6.
Worked solution
Each triangle has area magnitude 9 m. Displacement = 9 − 9 = 0 m, while distance = 9 + 9 = 18 m.
05 / Use only the requested time interval
The zero crossing may lie inside a restricted interval.
08 · Stop at five
For v = 6 − 2t on 0 ≤ t ≤ 5, find displacement and distance.
Hint
The negative triangle has width 2 and height 4.
Worked solution
Positive area = 9 m and negative magnitude = 4 m. Displacement = 5 m; distance = 13 m.
09 · Start at two
Find displacement and distance for the same model from t = 2 to t = 5.
Hint
Split at t = 3, where v = 0.
Worked solution
The positive triangle has area ½ × 1 × 2 = 1 m. The negative triangle has magnitude 4 m. Displacement = −3 m and distance = 5 m.
06 / An area is a change in position
Add the initial coordinate separately.
10 · Offset position
The particle with v = 6 − 2t starts at x = −4 m. Find its position at t = 5 s.
Hint
Its displacement over 0 to 5 is +5 m.
Worked solution
Final position = −4 + 5 = 1 m. The distance travelled is still 13 m; it does not depend on where the origin is chosen.
11 · Trapezium shortcut
Can ½(6 + (−6)) × 6 be used for the model on 0 to 6?
Hint
Which quantity does the signed calculation produce?
Worked solution
It gives the correct net displacement, 0 m. It does not give distance: split at the zero crossing and add positive magnitudes for that.
07 / Check units and what the graph represents
A speed graph has no negative values.
12 · Minutes on the axis
A velocity of 5 m/s is maintained for 2 minutes. Find displacement.
Hint
Convert 2 minutes to 120 seconds.
Worked solution
Displacement = 5 × 120 = 600 m. Multiplying 5 × 2 without conversion gives incompatible units.
13 · Speed graph
What does the area under a speed–time graph give?
Hint
Speed is the magnitude of velocity.
Worked solution
Total distance. Direction information has been removed, so that graph alone generally cannot give signed displacement.
14 · Curved graph
Can a triangle formula give the exact area under any curved velocity graph?
Hint
The shape must actually match the chosen geometry.
Worked solution
No. Curved boundaries generally require integration or a stated approximation. A straight chord is not automatically the true graph.
After an impact, use the velocity for each separate stage. A jump at a single instant contributes no finite area by itself; the intervals before and after still do.
08 / Choose signed sum or sum of magnitudes
Displacement, distance and final position are different outputs.
Mark zero crossings, calculate each region with consistent time units and then combine the regions for the requested quantity. Displacement adds signed areas. Distance adds their magnitudes. Final position adds the initial coordinate to displacement.
Section 1 of 8 · Area accumulates displacement