**Quantity on a time interval** | **Calculation** Displacement from t=a to t=b | s(b)-s(a)= _ab v(t), dt Distance travelled | _ab|v(t)|, dt Average velocity | s(b)-s(a)b-a Average speed | total distanceb-a Find the zeros of velocity, check its sign between them, and split the distance integral at direction changes. Example For v(t)=2t-2 on 0 <= t 3, the direction changes at t=1. Displacement= _03(2t-2), dt=3. Distance=- _01(2t-2), dt+ _13(2t-2), dt=1+4=5. The average velocity is 1, while the average speed is 53, in the appropriate units.

Find the zeros of velocity, check its sign between them, and split the distance integral at direction changes.

For \(v(t)=2t-2\) on \(0\le t\le3\), the direction changes at \(t=1\).

\(\text{Displacement}=\int_0^3(2t-2)\,\mathrm dt=3\).

\(\text{Distance}=-\int_0^1(2t-2)\,\mathrm dt+\int_1^3(2t-2)\,\mathrm dt=1+4=5\).

The average velocity is \(1\), while the average speed is \(\dfrac53\), in the appropriate units.

Kinematics

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Displacement and Total Distance

Displacement and Total Distance