Define displacement, velocity and acceleration, interpret motion graphs, and apply the equations of uniformly accelerated motion to one-dimensional problems

What this dot point is asking

SEAB wants you to define the kinematic quantities precisely, interpret displacement-time and velocity-time graphs, and apply the four equations of uniformly accelerated motion (often called the suvat equations) to one-dimensional problems, including vertical motion under gravity. This is the foundation of all dynamics and projectile work.

The answer

The defined quantities

• Displacement $s$: the vector change in position (units m).
• Velocity $v$: the rate of change of displacement, $v = \dfrac{\Delta s}{\Delta t}$ (units m s$^{-1}$).
• Acceleration $a$: the rate of change of velocity, $a = \dfrac{\Delta v}{\Delta t}$ (units m s$^{-2}$).

All three are vectors, so direction (sign in one dimension) matters throughout.

Reading motion graphs

The graphs encode the relationships geometrically:

• On a displacement-time graph, the gradient is the velocity.
• On a velocity-time graph, the gradient is the acceleration and the area under the line is the displacement.

A curved displacement-time graph means changing velocity; a sloped velocity-time line means constant acceleration.

The equations of uniformly accelerated motion

For constant acceleration $a$, with initial velocity $u$, final velocity $v$, displacement $s$ and time $t$:

Each equation omits one of the five variables. Choose the equation that contains the three quantities you know plus the one you want.

Vertical motion under gravity

Free fall is uniformly accelerated motion with $a = g$ (about $9.81\ \text{m s}^{-2}$) directed downward. Choose a sign convention (commonly upward positive) and apply it consistently: the launch speed, the acceleration and the displacement all carry signs relative to that choice.

Examples in context

Example 1. Stopping distance and road safety. A car at $20\ \text{m s}^{-1}$ braking at $5.0\ \text{m s}^{-2}$ needs $v^2 = u^2 + 2as \Rightarrow s = \dfrac{20^2}{2 \times 5.0} = 40\ \text{m}$ to stop. Doubling the speed to $40\ \text{m s}^{-1}$ quadruples the braking distance to $160\ \text{m}$, because $s \propto u^2$, a result with direct road-safety meaning.

Example 2. A dropped stone down a well. A stone dropped from rest falls for $2.0\ \text{s}$ before a splash is heard. Ignoring the sound travel time, the well depth is $s = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(2.0)^2 = 19.6\ \text{m}$. The quadratic dependence on time is why the second second of fall covers far more distance than the first.

Try this

Q1. Define velocity and acceleration, and state how each is found from a velocity-time graph. [3 marks]

• Cue. Velocity is rate of change of displacement; acceleration is rate of change of velocity (the gradient of the velocity-time graph). Displacement is the area under the velocity-time graph.

Q2. A cyclist accelerates from $4.0\ \text{m s}^{-1}$ to $10\ \text{m s}^{-1}$ in $3.0\ \text{s}$. Find the acceleration and the distance travelled. [3 marks]

• Cue. $a = \dfrac{10 - 4.0}{3.0} = 2.0\ \text{m s}^{-2}$; $s = \tfrac{1}{2}(4.0 + 10)(3.0) = 21\ \text{m}$.

Q3. An object is projected vertically upward and returns to its starting point. Sketch its velocity-time graph and explain what the area between the line and the time axis represents. [3 marks]

• Cue. A straight line of constant negative gradient crossing zero at the top; equal positive and negative areas, the net displacement being zero on return.