Average Velocity and Average Speed

Average quantities answer the examiner's favourite opening question: how fast, overall? — and they hide more traps per formula than any other topic in this chapter. Average velocity is displacement per unit time (a vector that inherits the sign of the displacement); average speed is total distance per unit time (a scalar that can never be negative). The two coincide only when the motion never reverses, and their relationship average speed ≥ |average velocity| is a ready-made Assertion–Reason item. Then comes the classic computational trap: for a journey made of two legs, the mean you must use depends on whether the legs matched in distance (harmonic mean) or in time (arithmetic mean) — the source of the famous 48 km/h answer that NEET has recycled for decades. This page develops all of it with derivations, the chord-slope graph skill, and verified numbers.

What is Average Velocity? — Complete Theory

1. Definitions and dimensional check

If a particle has displacement Δx = x₂ − x₁ in the time interval Δt = t₂ − t₁, the average velocity is

v̄ = Δx/Δt — a vector along the x-axis; SI unit m s⁻¹; dimensions [L T⁻¹].

The average speed over the same interval is

v̄_sp = total path length/Δt — a scalar; same unit and dimensions [L T⁻¹].

Dimensional check: [L]/[T] = [L T⁻¹] for both, as required — averaging cannot change the dimensions of the underlying quantity. Note what the definition of v̄ does not contain: any information about the route. A commuter who drives 10 km to office and 10 km back in one hour has average velocity zero and average speed 20 km h⁻¹ — both answers are correct simultaneously, because the two averages describe different aspects of the same motion. If you have not yet fixed the difference between distance and displacement, revise the distance and displacement page before continuing.

2. The comparison inequality and when equality holds

Since distance ≥ |displacement| for every motion, dividing both sides by the same positive Δt gives

average speed ≥ |average velocity|, with equality iff the direction never reverses during the interval.

Every reversal contributes extra path length that displacement cancels, so any turning widens the gap between the two averages. This is the direct analog of d ≥ |s| from the previous page, and examiners pair the two freely: "The average speed of a particle can be less than, equal to or greater than the magnitude of its average velocity — when?" Answer: it can only be ≥, with equality in unidirectional motion. Watch the sign too: average velocity can be negative (net motion along −x), while average speed cannot even be zero unless the particle never moves at all.

3. Chord slope: reading v̄ from a position-time graph

On an x–t graph, the average velocity between two times equals the slope of the straight chord joining the two points (t₁, x₁) and (t₂, x₂) — rise over run is literally Δx/Δt. This is the graphical form of the definition, and it works for any motion, curved or not. The tangent at a point, by contrast, gives the instantaneous velocity — the distinction is developed on the instantaneous velocity page with its own figure. For now, fix the pairing: chord ↔ average, tangent ↔ instantaneous. A chord sloping downward means negative average velocity; a horizontal chord means zero average velocity, even across an interval full of frantic motion.

4. The two-leg journey: harmonic vs arithmetic mean

When a journey splits into two legs at constant speeds v₁ and v₂, the correct average speed depends on what stayed constant:

Equal distances d each leg: total time = d/v₁ + d/v₂, so v̄_sp = 2d/(d/v₁ + d/v₂) = 2v₁v₂/(v₁ + v₂) — the harmonic mean. Derivation check: for v₁ = 40, v₂ = 60 → 2(2400)/100 = 48 km h⁻¹. The slow leg dominates because more time is spent in it — this is why the answer is always less than the arithmetic mean (50 here) and closer to the smaller speed.

Equal times t each leg: total distance = v₁t + v₂t, so v̄_sp = (v₁t + v₂t)/(2t) = (v₁ + v₂)/2 — the arithmetic mean.

Dimensional check on the harmonic mean: (m s⁻¹ · m s⁻¹)/(m s⁻¹) = m s⁻¹ ✔ — the formula is dimensionally a speed, a useful guard against the wrong-but-tempting (v₁ + v₂)/2 substitution in the equal-distance case.

Visualising Average Velocity: the Chord

Solved Examples on Average Velocity and Speed (Step-by-Step)

Solved Example 1 — The 48 km/h classic (equal distances)

Q. A car covers the first half of the distance between two towns at 40 km h⁻¹ and the second half at 60 km h⁻¹. Find its average speed.

Step 1. Legs hold distance equal → harmonic mean: v̄_sp = 2v₁v₂/(v₁ + v₂).

Step 2. Substitute: 2 × 40 × 60/(40 + 60) = 4800/100.

Step 3. v̄_sp = 48 km h⁻¹ — less than the arithmetic mean 50, pulled toward the slower leg where more time was spent.

Solved Example 2 — Average velocity from a position function

Q. A particle's position is x = 3t² + 2 (SI units). Find its average velocity between t = 0 and t = 2 s, and compare with the average speed.

Step 1. Positions: x(0) = 2 m; x(2) = 3(4) + 2 = 14 m.

Step 2. Displacement Δx = 14 − 2 = 12 m over Δt = 2 s → v̄ = 12/2 = +6 m s⁻¹.

Step 3. Is the path monotonic? v = dx/dt = 6t > 0 for t > 0 — no reversal, so distance = |displacement| and the average speed equals 6 m s⁻¹ too (equality case of the comparison inequality).

Practice Questions on Average Velocity and Speed (With Solutions)

Practice Q1 (Numerical-value type)

Q. A bus goes from stop A to stop B at 30 km h⁻¹ and immediately returns at 60 km h⁻¹. Its average speed for the round trip is ______ km h⁻¹.

Solution. Equal distances (A→B and B→A) → harmonic mean: 2 × 30 × 60/(30 + 60) = 3600/90 = 40 km h⁻¹. (Average velocity of the round trip is, of course, zero.)

Practice Q2 (Single-correct MCQ)

Q. Which statement is always true for motion along a line? (a) average speed ≤ |average velocity| (b) average speed ≥ |average velocity| (c) average speed = |average velocity| always (d) average velocity is never zero

Solution. Dividing d ≥ |Δx| by Δt > 0 gives v̄_sp ≥ |v̄|, with equality only without reversal — so (b) is the always-true statement; (c) fails for round trips and (d) fails for any closed trip.

Practice Q3 (Assertion–Reason)

Q. Assertion (A): A man walks 3 km east and then 4 km west in one hour; his average velocity is 1 km h⁻¹ west. Reason (R): Average velocity is displacement divided by total time.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true

Solution. Taking east positive: Δx = 3 − 4 = −1 km (1 km west) in 1 h → v̄ = −1 km h⁻¹, i.e. 1 km h⁻¹ west ✔. R is the definition that produced A.

Key Formulas & Takeaways

FAQs on Average Velocity and Average Speed

What is the difference between average velocity and average speed?

Average velocity is displacement divided by time — a vector carrying the sign of the displacement. Average speed is total distance divided by time — a scalar that is never negative.

When is average speed the arithmetic mean of two speeds?

Only when the two speeds are maintained for equal time intervals, giving (v₁ + v₂)/2. If the distances covered are equal instead, the average speed is the harmonic mean 2v₁v₂/(v₁ + v₂).

A car travels at 40 km/h and 60 km/h over equal distances. What is its average speed?

The average speed is 2 × 40 × 60/(40 + 60) = 48 km/h — the harmonic mean, which is always less than the arithmetic mean of 50 km/h for this case.

Can average velocity be zero while average speed is not?

Yes — any out-and-back trip has zero net displacement, so the average velocity is zero, while the average speed stays positive because distance is always travelled.