Position, velocity, acceleration

Motion is tracked by three quantities built from one idea: how position changes over time.

Displacement vs distance

Position is where you are relative to an origin. Displacement is the change in position, a straight-line quantity with a sign, while distance is the total path length traveled and is always positive. Walk 40 m east then 30 m west and your distance is 70 m but your displacement is only 10 m east.

\Delta x = x_f - x_i

Velocity and graphs

Average velocity is displacement divided by elapsed time; instantaneous velocity is what that quotient becomes over a vanishing interval, which is the slope of the position-time graph. Acceleration is the rate of change of velocity, the slope of the velocity-time graph. On a velocity-time graph, the area under the curve equals the displacement.

v_{\text{avg}} = \frac{\Delta x}{\Delta t}, \quad a = \frac{\Delta v}{\Delta t}

Worked example

Let s(t) = t^2 - 4t + 3 give position in meters. From t = 0 to t = 3, displacement is s(3) - s(0) = 0 - 3 = -3 m, so average velocity is -3/3 = -1 m/s even though the object ends up moving forward: v(t) = 2t - 4 gives +2 m/s at t = 3. The object reverses at t = 2 (position falls from 3 m to -1 m, then rises back to 0 m), so the distance traveled is 4 + 1 = 5 m while displacement is only -3 m.

v(3) = 2(3) - 4 = 2 \text{ m/s}, \quad \text{distance} = 5 \text{ m}

Flashcard

v_{\text{avg}} = \frac{\Delta x}{\Delta t}

Average velocity is displacement over elapsed time

This is the slope of the secant line on a position-time graph between two instants. It can be zero even when a lot of ground was covered, if the motion ends where it started.

Check your understanding

A cyclist's position is s(t) = 2t^2 - 3t meters after t seconds. What is her average velocity from t = 1 to t = 2?

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