Displacement
Motion & Kinematics - OCR A-Level Physics
Key Definition
Displacement
The straight-line distance from a fixed reference point in a specified direction. It is a vectorA quantity with both magnitude and direction. quantity, measured in metres ($\text{m}$). Symbol $s$.
The straight-line distance from a fixed reference point in a specified direction. It is a vectorA quantity with both magnitude and direction. quantity, measured in metres ($\text{m}$). Symbol $s$.
Distance vs Displacement
Distance is the total length of path travelled, a scalarA quantity with magnitude only, no direction. measured in $\text{m}$. Displacement is the straight-line vector from start to end. A runner finishing one lap of a 400 $\text{m}$ track has travelled a distance of $400 \text{ m}$ but a displacement of $0 \text{ m}$.
Velocity
Velocity
The rate of change of displacement. A vector measured in $\text{m s}^{-1}$. Symbol $v$.
The rate of change of displacement. A vector measured in $\text{m s}^{-1}$. Symbol $v$.
$$v = \frac{\Delta s}{\Delta t}$$
Speed
Speed is the magnitude of velocity, a scalar in $\text{m s}^{-1}$. A car driving at $30 \text{ m s}^{-1}$ round a bend has constant speed but changing velocity, because its direction is changing.
Acceleration
Acceleration
The rate of change of velocity. A vector measured in $\text{m s}^{-2}$. Symbol $a$.
The rate of change of velocity. A vector measured in $\text{m s}^{-2}$. Symbol $a$.
$$a = \frac{\Delta v}{\Delta t}$$
Instantaneous vs average
Instantaneous velocity is the velocity at one moment, found from the gradient of a displacement-time graph at that point. Average velocity is the total displacement divided by total time: $\bar{v} = \frac{\Delta s}{\Delta t}$.
- A car travels $500 \text{ m}$ north then $500 \text{ m}$ east in $50 \text{ s}$. Distance $= 1000 \text{ m}$, displacement $= \sqrt{500^2 + 500^2} = 707 \text{ m}$ at $45^\circ$ east of north.
- Average speed of that car $= \frac{1000}{50} = 20 \text{ m s}^{-1}$. Average velocity $= \frac{707}{50} = 14.1 \text{ m s}^{-1}$ at $45^\circ$ east of north.
- A sprinter accelerating from rest to $10 \text{ m s}^{-1}$ in $2 \text{ s}$ has $a = \frac{10 - 0}{2} = 5 \text{ m s}^{-2}$.
- Deceleration is just negative acceleration. A car braking from $20 \text{ m s}^{-1}$ to rest in $4 \text{ s}$ has $a = -5 \text{ m s}^{-2}$.
Common Mistake
MEDIUM
Wrong: Treating displacement and distance as interchangeable, or quoting velocity as a single number with no direction stated.
Right: Displacement is a vector (state direction or sign); distance is a scalar. For motion in a straight line, choose a positive direction and use signed values. A runner returning to the start has zero displacement but non-zero distance.
Right: Displacement is a vector (state direction or sign); distance is a scalar. For motion in a straight line, choose a positive direction and use signed values. A runner returning to the start has zero displacement but non-zero distance.
Examiner Tips and Tricks
- When a question asks for "the velocity", you must state both a magnitude and a direction (or a sign).
- If the question gives a speed and direction separately, state both in the answer to score the vector mark.
- "Decelerates uniformly" tells you acceleration is constant and negative. List $a$ with a minus sign before using SUVAT.