define and use distance, displacement, speed, velocity and acceleration

Equations of Motion

Distance, Displacement, Speed, Velocity and Acceleration

Distance is a scalar quantity that measures the total length of the path travelled. It is always positive and has units of metres (m).

Displacement is a vector quantity that represents the change in position of an object. It is defined as the straight‑line vector from the initial position to the final position and can be positive, negative or zero.

Speed is the rate of change of distance with respect to time. It is a scalar and is given by

$$v_{\text{avg}} = \frac{\text{distance}}{\Delta t}$$

Velocity is the rate of change of displacement with respect to time. It is a vector and is given by

$$\vec{v}_{\text{avg}} = \frac{\Delta \vec{s}}{\Delta t}$$

Acceleration is the rate of change of velocity with respect to time. It is a vector and is given by

$$\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$$

Key Relationships

• Speed is the magnitude of velocity: $v = |\vec{v}|$.
• Acceleration is the derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt}$.
• Velocity is the derivative of displacement: $\vec{v} = \frac{d\vec{s}}{dt}$.
• Speed is the derivative of distance: $v = \frac{ds}{dt}$.

Kinematic Equations for Constant Acceleration

Example Problem

A car starts from rest and accelerates uniformly at $2.0\,\text{m s}^{-2}$ for $10\,\text{s}$. Calculate:

1. The final velocity.
2. The distance travelled.
3. The average speed.

Solution:

1. Using $\displaystyle \vec{v} = \vec{v}_0 + \vec{a}t$ with $\vec{v}_0 = 0$: $$\vec{v} = 0 + (2.0\,\text{m s}^{-2})(10\,\text{s}) = 20.0\,\text{m s}^{-1}.$$
2. Using $\displaystyle \Delta \vec{s} = \vec{v}_0 t + \tfrac{1}{2}\vec{a}t^2$: $$\Delta \vec{s} = 0 + \tfrac{1}{2}(2.0\,\text{m s}^{-2})(10\,\text{s})^2 = 100\,\text{m}.$$
3. Average speed equals total distance divided by total time: $$v_{\text{avg}} = \frac{100\,\text{m}}{10\,\text{s}} = 10.0\,\text{m s}^{-1}.$$

Suggested Diagram