01 · Reversal
A trolley moves 11 m right and then 4 m left. Find distance and displacement.
Hint
Choose right positive.
Worked solution
Distance=11+4=15 m. Displacement=11−4=7 m right.
Understand · explore · practise
Distance and displacement on a journey
Add vector changes for net displacement and straight-leg lengths for total distance, including return trips and average velocity.
Before you startVector components and magnitudes; Pythagoras.
01 / The endpoint does not determine the whole journey
Distance records the path; displacement records the net change.
Add vectors for displacement; add travelled lengths for distance.
For straight legs with displacement vectors a and b, net displacement is a+b, while distance is |a|+|b|. This is generally different from |a+b|.
Compare the route with the net arrowExplore
Each listed leg is straight. Coordinates and lengths are in metres, with right and up positive. Orange follows the route; blue joins start to finish.
02 / On a line, use signed changes and absolute lengths
A return leg subtracts from displacement but adds to distance.
02 · Back to start
It then moves another 7 m left. Find total distance and final displacement.
Hint
Include the final leg in both totals.
Worked solution
Distance=11+4+7=22 m. Net displacement=11−4−7=0 m.
03 · Three stages
A walker moves 5 m west, 2 m east and 6 m west. Find total distance and displacement.
Hint
Keep one direction convention throughout.
Worked solution
Distance=5+2+6=13 m. With east positive, displacement=−5+2−6=−9 m, or 9 m west.
03 / Add corresponding components
Pythagoras applies to each straight leg when finding distance.
Two straight displacements are a=3i+4j m and b=3i−4j m.Worked example
a+b=6i m
The j-components cancel.
|a|=|b|=5 m
Each leg is a 3–4–5 triangle.
Distance=10 m; displacement magnitude=6 m
The bent route is longer than the straight connection.
Watch: two legs and their direct displacement
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04 · Another pair
Two straight legs have displacements 4i+3j m and −4i+3j m. Find net displacement and distance.
Hint
Add components, then add individual magnitudes.
Worked solution
Net displacement=6j m with magnitude 6 m. Each leg has length 5 m, so distance=10 m.
05 · Square route
Three consecutive sides of a square of side 4 m are travelled. Find distance and displacement magnitude.
Hint
The direct connection is the missing fourth side.
Worked solution
Distance=12 m; displacement magnitude=4 m.
04 / Position subtraction gives the net vector
It gives no record of detours.
06 · Coordinates
A journey starts at (−2,1) m and ends at (4,9) m. Find net displacement and its magnitude.
Hint
Final minus initial.
Worked solution
Displacement=(6,8) m, magnitude √(36+64)=10 m.
07 · Unknown route
Can its distance be 8 m? Can it be 14 m?
Hint
Compare with the straight-line separation.
Worked solution
It cannot be 8 m, which is shorter than the 10 m separation. It can be 14 m, for example a straight 6 m horizontal leg followed by an 8 m vertical leg.
08 · Minimum
What is the minimum possible distance between those endpoints?
Hint
Use the direct route.
Worked solution
10 m. A direct straight path without reversal achieves it; stationary pauses would change time but not distance.
05 / Average speed and average velocity use different numerators
Use the same total elapsed time in both.
A journey follows the model route 6 m right then 8 m up in 20 s.
09 · Average speed
Find the average speed.
Hint
Use total distance.
Worked solution
(6+8)/20=0.7 m/s.
10 · Average velocity
Find average velocity as a vector and its magnitude.
Hint
Use net displacement divided by time.
Worked solution
Average velocity=(6i+8j)/20=0.3i+0.4j m/s. Its magnitude is 10/20=0.5 m/s, different from the average speed.
11 · Return-trip average
Why can a complete return trip have zero average velocity but positive average speed?
Hint
Its endpoint and path totals differ.
Worked solution
Net displacement is zero, so average velocity is zero. A positive travelled distance divided by positive elapsed time gives positive average speed.
06 / The direct connection is no longer than the route
Equality has a geometric condition.
12 · General comparison
State the relationship between distance travelled D and net displacement magnitude R.
Hint
Consider the shortest connection.
Worked solution
D≥R. For straight segments, equality requires the nonzero displacements to lie along the same direction without reversal.
13 · Curved leg
If one leg is curved, can you replace its travelled length by the magnitude of its endpoint displacement?
Hint
The straight chord may shorten the path.
Worked solution
Not generally. The magnitude gives the chord length; a curved path can be longer. Add actual path lengths when calculating distance.
07 / Keep vector units and direction consistent
Metres do not turn into kilometres after taking a square root.
14 · Units
If displacement components are 6 m and 8 m, explain why the magnitude is 10 m rather than 10 km or 10 m².
Hint
Square, add and then square-root the units.
Worked solution
The sum of squares has units m²; taking its square root gives metres. No kilometre conversion has occurred.
A diagram should state its axes and scale. Check every leg direction before adding; use a separate distance total so a reversal is not accidentally treated as negative travelled length.
08 / Calculate two different totals
Use the question’s wording to choose the answer.
Net displacement is the vector sum or final minus initial position. Total distance adds the lengths actually travelled. Average velocity uses net displacement; average speed uses total distance. The direct separation is a lower bound on path length, not a substitute for it.
Section 1 of 8 · The endpoint does not determine the whole journey