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Lesson 2 of 10

Motion in a Straight Line · Lesson 2 of 10

Average Velocity and Average Speed

“Use total displacement and total time correctly, including return journeys, unequal intervals and rest periods.”

Learning Objectives

• Calculate average velocity and average speed for a complete interval. • Explain why average speed is at least the magnitude of average velocity. • Combine journeys using time weighting rather than blindly averaging speeds. • Interpret the secant slope of a position-time graph.

A rate for a whole interval

Suppose a car moves 120 m east in 10 s. Its average velocity is +12 m/s when east is positive. This says that the net change in position is 12 metres per second averaged over that interval. It does not say the car moved at 12 m/s at every instant. Different patterns of speeding up, slowing down and resting can share the same average velocity.

Average velocityLaTeX
Use elapsed time Δt > 0. The sign is the sign of displacement. SI unit m s⁻¹; dimension [L T⁻¹].
Average speedLaTeX
D is total distance travelled, including travel in both directions. Include all time in the stated interval, including stops.

Because D is at least |Δx|, division by the same positive elapsed time gives average speed ≥ |average velocity|. They are equal if the object never reverses during the interval. A complete round trip has zero average velocity but positive average speed if any distance was covered. Both are zero only if no distance was covered.

The same trip, two averages

Problem
A cyclist goes 120 m east in 10 s, then 40 m west in 5 s. Find both averages.

  1. 1.Elapsed time is 10 + 5 = 15 s. The displacement is 120 − 40 = +80 m.
  2. 2.Average velocity is 80/15 = +5.33 m/s.
  3. 3.Distance is 120 + 40 = 160 m, so average speed is 160/15 = 10.67 m/s. Averaging the two speeds, 12 and 8 m/s, would give 10 m/s and is wrong because the durations differ.

Which interval are you averaging?

An average is incomplete without its time interval. During part of a journey, endpoints may be far apart; over a longer interval the traveller may return near the start. Always first identify the two times, then determine the positions and distance travelled between them. Do not automatically use the full journey if the question stops halfway through the return.

Walking to a closed market

Problem
A market is 2.5 km from home. A person walks there at 5 km/h, immediately turns back, and returns at 7.5 km/h. Compare the first 30, 40 and 50 minutes.

  1. 1.The outward time is 2.5/5 = 0.5 h = 30 min. The return time is 2.5/7.5 = 1/3 h = 20 min.
  2. 2.From 0 to 30 min: displacement and distance are both 2.5 km, so average velocity is +5 km/h and average speed is 5 km/h.
  3. 3.From 0 to 40 min: the person has returned for 1/6 h, travelling 1.25 km back. Displacement is +1.25 km, distance is 3.75 km and elapsed time is 2/3 h. Average velocity is +1.875 km/h and average speed is 5.625 km/h.
  4. 4.From 0 to 50 min: displacement is zero, distance is 5 km and elapsed time is 5/6 h. Average velocity is zero and average speed is 6 km/h. The zero describes net progress, not lack of effort.

Why simple averaging often fails

Time-weighted average velocityLaTeX
Each v_i is the signed constant velocity, or the average velocity, on that segment. Use speeds in the numerator instead when calculating total distance per time.

For equal durations, this becomes (v₁ + v₂)/2. For unequal durations, the longer-lasting velocity has greater weight. For equal distances travelled at positive speeds u and w, the times are d/u and d/w. Total distance 2d divided by their sum gives 2uw/(u + w), the harmonic mean. The arithmetic mean is generally too high in this equal-distance case because the slower leg takes longer.

Equal distances at different speeds

Problem
Travel equal distances at 30 km/h and 60 km/h. Find average speed.

  1. 1.Take each leg as 60 km. The times are 2 h and 1 h.
  2. 2.Average speed = 120 km / 3 h = 40 km/h, not 45 km/h.
  3. 3.If instead each speed lasted one hour, the distances would be 30 and 60 km and the average speed would be 45 km/h. The condition about equal time or equal distance is essential.

Stops count in the elapsed time

If you travel 600 m in 5 min and then wait for 5 min, the average speed over the full 10 min is 600/600 = 1 m/s. Over only the moving portion it is 600/300 = 2 m/s. Neither is inherently wrong; they answer different questions. Write down the interval to avoid silently removing rest time.

Average velocity on a graph

Join the initial and final points of a position-time curve with a straight chord, called a secant. Its vertical rise is displacement, its horizontal run is elapsed time, and their ratio is average velocity. Intermediate bends affect distance travelled but not this secant slope. A horizontal secant can connect the endpoints of a non-horizontal curve: the object has returned to its starting position.

Quiz

Quick check

For a complete round trip with nonzero distance travelled, which statement is true?

Key Takeaways

Key Takeaways

• Always attach an average to a specified time interval. • Average velocity uses displacement; average speed uses total path length. • Count rest time when it belongs to the interval. • Combine segments through total distance or displacement and total time.

Previous · Lesson 1

Position, Displacement and the Point-Object Model

Next · Lesson 3

Instantaneous Velocity and Speed