01 / CHAPTER 02
Displacement and average velocity
Separate where an object is from how far it traveled and how fast the change occurred.
ANCHOR EQUATION
Displacement
In symbols: Δx = x₂ − x₁
Displacement is a signed change in position, not the total distance traveled.
Displacement keeps the sign
Distance answers ‘how much path was covered’; displacement answers ‘where did the final position land relative to the initial position?’ Only displacement can add consistently across segments because it retains direction.
Subtract positions, do not add path lengths
Let an object begin at x₁ and finish at x₂. The signed change is Δx = x₂ − x₁. If a trip has segments Δx₁, Δx₂, …, the total displacement is their sum because intermediate positions cancel: (x₂ − x₁) + (x₃ − x₂) = x₃ − x₁. Distance instead sums absolute path lengths, |Δx₁| + |Δx₂| + … .
A runner goes from x = 0 to 80 m, then back to x = 30 m. Distance = 80 + 50 = 130 m; displacement = 30 − 0 = +30 m. The average velocity over 26 s is 30/26 = 1.15 m/s, while average speed is 130/26 = 5.0 m/s.
The return-trip test
If the runner returns to the starting point, Δx = 0 and average velocity is zero even though distance and average speed are nonzero. This is why a GPS trace cannot infer velocity from total distance alone.
Work it: A cart moves −12 m, then +7 m, then +9 m in 8.0 s. Compute distance, displacement, average speed, and average velocity; state which quantities change sign if the positive direction is reversed.