01 · Distance
Find total distance for all four stages.
Hint
Add nonnegative travelled lengths.
Worked solution
6+0+9+3=18km.
Understand · explore · practise
Average speed and velocity from graphs
Calculate whole-journey average speed and velocity from displacement graphs, including waiting, reversals, nonzero starting positions and mixed time units.
Before you startReading displacement–time graphs; distance versus displacement.
01 / Keep two totals and one elapsed time
A return to the start does not erase the distance travelled.
Average velocity = net displacement / elapsed time. Average speed = total distance / elapsed time.
For a straight-line journey represented by straight position-time segments, add absolute position changes for distance and signed changes for displacement. Include waiting time in a whole-journey average.
Include the waiting timeExplore
Position is in km east of a fixed origin; times are minutes after departure. Straight segments join (0,2), (20,8), (35,8), (65,−1), (80,2). Select an endpoint to analyse the journey from t=0 to that time.
02 / Build a stage table before averaging
Changing direction affects displacement and distance differently.
Use the complete model journey.Worked example
First stage: +6 km in20min
Distance6km, signed displacement+6km.
Wait: 0 km in15min
Both displacement and distance changes are zero, but time increases.
Third stage: −9 km in30min
Distance9km, displacement−9km.
Last stage: +3 km in15min
Distance3km, displacement+3km.
02 · Displacement
Find net displacement and compare it with final position.
Hint
Subtract starting position from final position.
Worked solution
Net displacement=2−2=0km. The final position is2km east of the origin, which is not zero.
03 / Convert the full duration consistently
Minutes on the axis do not directly give km/h.
03 · Full duration
Express80min in hours and calculate whole-journey average speed.
Hint
Divide minutes by60.
Worked solution
80min=4/3h. Average speed=18/(4/3)=13.5km/h.
04 · Average velocity
Calculate the whole-journey average velocity.
Hint
Use net displacement, not18km.
Worked solution
0/(4/3)=0km/h. This does not imply rest during the journey.
05 · Segment velocity
Find the velocity from35 to65min in km/h.
Hint
The signed change is−9km in half an hour.
Worked solution
Velocity=−9/0.5=−18km/h, meaning18km/h westward.
04 / A pause changes a whole-journey average
It changes elapsed time even when position stays fixed.
Watch: the same distance over a longer time
Pause, replay or seek freely. The notes explain the same idea and stay in view.
06 · Before waiting
Find average speed from0 to20min.
Hint
Distance6km; time1/3h.
Worked solution
6/(1/3)=18km/h.
07 · After waiting
Find average speed from0 to35min.
Hint
The distance remains6km.
Worked solution
6/(35/60)=72/7km/h, about10.29km/h. The wait reduces the whole-period average.
08 · Moving-only average
Find average speed during moving time only for the full journey. Explain why it differs from13.5km/h.
Hint
Remove only the15-minute stop from the denominator.
Worked solution
Moving time = 65 min = 13/12 h. Moving-only average speed = 18 ÷ (13/12) = 216/13 ≈ 16.62 km/h. It answers a different question from the average over all80min.
05 / A negative coordinate is not negative distance
Subtract coordinates before dividing by elapsed time.
09 · Intermediate endpoint
Find net displacement fromt=0 tot=65min.
Hint
The initial position is2km; final position is−1km.
Worked solution
Net displacement=−1−2=−3km, or3km west.
10 · Two averages
Find average velocity and average speed over those65min.
Hint
Distance is6+9=15km.
Worked solution
Average velocity = −3 ÷ (65/60) = −36/13 ≈ −2.77 km/h. Average speed = 15 ÷ (65/60) = 180/13 ≈ 13.85 km/h.
06 / Do not average stage speeds without checking durations
The time spent at each speed matters.
11 · Unequal durations
A cyclist rides at12km/h for30min and24km/h for15min in one direction. Find average speed.
Hint
Calculate each distance first.
Worked solution
Distances6km and6km give12km in0.75h, so average speed16km/h. The unweighted mean18km/h is wrong because durations differ.
12 · Equal durations
A cyclist rides at12km/h for20min and24km/h for20min. Find average speed.
Hint
Now equal-time weighting is justified.
Worked solution
Distance=4+8=12km; time2/3h, giving18km/h. Here the ordinary mean works because the durations are equal.
07 / Use enough graph information to justify distance
Endpoint displacement alone may hide a reversal.
13 · Hidden motion
A curved position-time graph starts and ends at2m. Can you infer zero distance travelled?
Hint
Only endpoint displacement is known.
Worked solution
No. Net displacement is zero, but the curve may rise and fall. Split at each turning point and add absolute changes in position, provided the motion between turning points is monotonic.
14 · Comparison
For positive elapsed time, compare average speed with the magnitude of average velocity.
Hint
Use distance≥displacement magnitude.
Worked solution
Average speed is at least the magnitude of average velocity. Equality holds for a direct one-direction journey without reversal, even if it includes stops.
On a graph whose axes show kilometres and minutes, slopes initially have units km/min. Convert these carefully, or convert the time first. Do not measure the drawn length of a sloping graph line as physical distance travelled.
08 / Use the requested interval from start to finish
Include stops unless the question explicitly asks for moving time.
Net displacement is final minus initial position. Distance adds absolute changes between reversals. Divide the appropriate total by the same elapsed time, with consistent units. Stage speeds require time weighting; a return journey can have zero average velocity and positive average speed.
Section 1 of 8 · Keep two totals and one elapsed time