Module 3: Kinematics · Worked Examples · Prerequisites: K1, K3
Four journeys, and the question that decides every one of them: what is being accumulated, and what is being divided by?
Mahavakya
An average compresses a history into a single number. What survives the compression depends entirely on what you divided by.
Averages are computed in one step, which is why they usually appear as one-line questions. These four do not. Each takes a single journey and asks several things of it, because the interest lies in how the answers differ.
The four
1. One journey, seven questions
Problem
A cyclist rides 12 km north in 30 min, rests for 10 min, rides 4 km south in 15 min, then 3 km east in 15 min.
Find (a) the total distance (b) the displacement (c) the average speed (d) the average velocity (e) whether s ≥ |Δr| holds (f) the average speed while moving (g) which of (c) and (f) a cyclometer would show.
(a) Distance — add the legs, ignoring direction:
s = 12 + 4 + 3 = 19 km
(b) Displacement — net north is 12 − 4 = 8 km, and east is 3 km:
Δr = 3 i + 8 j km
|Δr| = √(9 + 64) = √73 ≈ 8.54 km
(c) Average speed. Total time is 30 + 10 + 15 + 15 = 70 min = 7/6 h:
19 ÷ (7/6) = 114/7 ≈ 16.3 km/h
(d) Average velocity — displacement over the same total time:
√73 ÷ (7/6) ≈ 7.32 km/h, directed 20.6° east of north
(e) 19 ≥ 8.54 ✓. The inequality is strict, as it must be — the journey reversed and then turned, so no two legs point the same way.
(f) Moving time is 60 min = 1 h, so the average while moving is 19 ÷ 1 = 19 km/h.
(g) A cyclometer shows 19 km/h — the moving average. It stops counting when the wheel stops, so the rest is absent from its denominator but not from yours.
This is the part students cannot do, and it is the whole point. Three different numbers — 16.3, 19 and 7.32 — all correctly describe the same ride. Which one is right depends on what is being accumulated and what it is divided by, and no amount of care with the arithmetic decides that for you.
2. Two cars, same speeds, different answers
Problem
Car A covers the first half of the distance at 30 km/h and the second half at 60 km/h. Car B spends the first half of the time at 30 km/h and the second half at 60 km/h.
(a) Find each average speed. (b) Say which is larger before calculating, and why. (c) Show the equal-distance average is always the smaller. (d) When are they equal?
(b) first, since it costs nothing. Car A spends longer at the slow speed, because covering the same distance at 30 takes twice as long as at 60. Car B splits its time evenly. So A’s average is dragged toward 30, and B’s average must be larger.
(a) Car A — take each half as d:
2d ÷ (d/30 + d/60) = 2(30)(60)/(30 + 60)
= 40 km/h — the harmonic mean
Car B — take each half as t:
(30t + 60t) ÷ 2t = (30 + 60)/2
= 45 km/h — the arithmetic mean
(c) Subtracting the two general expressions:
(v1 + v2)/2 − 2v1v2/(v1 + v2) = (v1 − v2)² / 2(v1 + v2)
A square divided by something positive — so the difference is never negative.
With 30 and 60 that difference is (30)²/180 = 5 km/h, which matches 45 − 40.
(d) The difference vanishes only when v1 = v2 — the degenerate case where the journey has one speed and the question does not arise.
Neither 40 nor 45 is “the average of 30 and 60”. Both are correct averages of the same two speeds, for two different journeys. The phrase “the first half” is what carries the ambiguity, and a problem that does not say half of what has not been fully stated.
3. A journey that reverses
Problem
A particle moves with v = 8 − 2t m/s for 0 ≤ t ≤ 6 s.
Find (a) when it reverses (b) its displacement (c) the distance travelled (d) both averages (e) why they differ.
(a) v = 0 at t = 4 s, and the sign changes there — positive before, negative after — so it is a genuine turning point.
(b) Displacement is the signed area under v–t, or equivalently ∫v dt:
Δx = 8(6) − 6² = 48 − 36 = +12 m
(c) Distance must be split at the reversal:
0 to 4 s: 8(4) − 4² = +16 m
4 to 6 s: 12 − 16 = −4 m
s = 16 + 4 = 20 m
(d) Average velocity = 12/6 = 2 m/s. Average speed = 20/6 = 3.33 m/s.
(e) The four metres travelled backward are counted twice over. They cancel from the displacement — reducing it from 16 to 12 — and they add to the distance, raising it from 16 to 20. That single stretch of track is the entire difference between 2 and 3.33 m/s, and every reversal in every problem does exactly this.
4. Three legs, two criteria
Problem
A journey is divided into three equal distances, covered at 20, 30 and 60 km/h.
(a) Find the average speed. (b) Find it again if the three legs were equal in time instead. (c) Comment on the numbers.
(a) Equal distances. Take each leg as d. The times are d/20, d/30 and d/60:
3d ÷ d(1/20 + 1/30 + 1/60) = 3 ÷ (3/60 + 2/60 + 1/60)
= 3 ÷ (6/60) = 30 km/h — the harmonic mean of three
(b) Equal times. Take each as t:
(20 + 30 + 60)/3 = 110/3 ≈ 36.7 km/h
(c) The equal-distance answer is exactly 30 km/h — the middle speed, which looks like a coincidence and is one. It happens because the three reciprocals sum to 6/60, and nothing about the middle value causes it.
The equal-time answer is larger, as it must be. The pattern from example 2 holds for any number of legs: equal distances always give the smaller average, because equal distances mean more time spent at the lower speeds, and time is what the average is divided by.
What the four have in common
Not one of them needed a formula for the harmonic mean. Every answer came from the same two lines: find the total distance, find the total time, divide.
The formulas exist, and they are worth recognising when they appear — but a student who reconstructs them from the definition will never apply the arithmetic mean to a distance-split journey, and a student who remembers two formulas will do it at least once.
Method, in three lines
- Ask what is being accumulated — distance, or displacement.
- Ask what it is divided by — total time, or moving time.
- Assemble both totals, then divide. The formulas are what that produces.
Where these sit
Examples 1 to 3 are JEE Main standard. Example 4 is Advanced for its part (c), which asks for a general statement rather than a number. All four assume K3; example 1 also uses K1’s distinction between distance and displacement, and example 3 anticipates K6’s signed area.
Prerequisites: K1 and K3 · Working time: 30 min · Level: JEE Main to Advanced