3.4.1.3
Displacement, velocity and acceleration are the vector forms of distance, speed and time-rate quantities
Motion Along a Straight Line — AQA A-Level Physics
Key Definition
Displacement — The distance of an object from a fixed point in a specified direction. Symbol: s. Unit: m.
Key Definition
Velocity — The rate of change of displacement. Symbol: v. Unit: m s^-1.
$$v = \frac{\Delta s}{\Delta t}$$
- Where:
- $v$ = velocity (m s⁻¹)
- $Δs$ = change in displacement (m)
- $Δt$ = change in time (s)
Key Definition
Acceleration — The rate of change of velocity. Symbol: a. Unit: m s^-2.
$$a = \frac{\Delta v}{\Delta t}$$
- Where:
- $a$ = acceleration (m s⁻²)
- $Δv$ = change in velocity (m s⁻¹)
- $Δt$ = change in time (s)
$$v = \frac{\Delta s}{\Delta t}$$
- $v$: velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. (m s⁻¹)
- $Δs$: change in displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m)
- $Δt$: change in time (s)
$$a = \frac{\Delta v}{\Delta t}$$
- $a$: accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². (m s⁻²)
- $Δv$: change in velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. (m s⁻¹)
- $Δt$: change in time (s)
Worked Example
A car accelerates uniformly from rest to 150 km h^-1 in 6.2 s. Calculate the accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻². in m s^-2.
Show Solution
1
Convert speed to m s^-1
$$150 \text{ km h}^{-1} = \frac{150 \times 10^3}{3600} = 41.67 \text{ m s}^{-1}$$
2
Apply the accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻². equation
$$a = \frac{\Delta v}{\Delta t} = \frac{41.67}{6.2} = 6.7 \text{ m s}^{-2}$$
Answer
$a = 6.7$ m s$^{-2}$