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Physics · Ch 2 — Motion in a Straight Line

Average Speed and Average Velocity

2.3

Average Speed and Average Velocity

Once we can measure how far an object has travelled (path length) and by how much its position has changed (displacement) over an interval of time, we can define two related — but generally different — measures of "how fast" the object moved on average during that interval.

Average speed. The average speed of an object over a time interval Δt=t2−t1\Delta t = t_2 - t_1 is defined as the total path length covered divided by the time taken:

average speed=total path lengthΔt\text{average speed} = \frac{\text{total path length}}{\Delta t}

Average speed is a scalar quantity — it is always positive and carries no direction. It tells us how much ground, in total, the object covered per unit time, without regard to whether that ground was covered going forward, backward, or both.

Average velocity. The average velocity of an object over the same interval is defined as its displacement divided by the time taken:

vˉ=ΔxΔt=x2−x1t2−t1\bar v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}

Average velocity is a vector quantity along the line of motion — its sign indicates the net direction of motion over the interval, and its magnitude tells us the net rate of change of position, ignoring any reversals along the way.

Why average speed can differ from (and never be less than) the magnitude of average velocity. Since path length is always at least as large as the magnitude of displacement over the same interval (Section 2.2), dividing both by the same time interval Δt\Delta t gives

average speed≥∣vˉ∣\text{average speed} \ge |\bar v|

Equality holds if and only if the object never changes direction during the interval — in that special case the path length and the magnitude of the displacement are identical.

Worked illustration. Suppose a car travels 400 m due east in 20 s and then reverses and travels 150 m due west in the next 10 s. Its total path length is 400+150=550400 + 150 = 550 m over a total time of 3030 s, giving an average speed of 550/30≈18.3550/30 \approx 18.3 m/s. Its net displacement, however, is 400−150=250400 - 150 = 250 m east, giving an average velocity of 250/30≈8.3250/30 \approx 8.3 m/s east — considerably smaller in magnitude than the average speed, precisely because the car reversed direction partway through the journey. (This example is worked out in full in Example 2.) …

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