Displacement, velocity and acceleration, interpreting motion graphs, the equations of motion for constant acceleration, and projectile motion.

What this dot point is asking

WJEC wants you to define displacement, velocity and acceleration, read information from motion graphs, apply the four equations of motion for constant acceleration, and solve projectile problems by treating horizontal and vertical motion independently. This is the foundation of the whole mechanics strand, so the examiners test it both as standalone calculation questions and as the opening step of longer dynamics and energy problems.

The answer

Defining the quantities

Displacement is the vector distance from a defined start point, measured in metres. Velocity is the rate of change of displacement, a vector. Acceleration is the rate of change of velocity, also a vector.

Distance and speed are the scalar partners of displacement and velocity. WJEC expects you to distinguish them: a runner completing one lap of a $400\,\text{m}$ track has travelled a distance of $400\,\text{m}$ but has a displacement of zero.

Motion graphs

A curved displacement-time graph means changing velocity, so you take the gradient of a tangent at a point to find the instantaneous velocity. On a velocity-time graph, splitting the area into triangles and rectangles is the quickest way to get displacement when the acceleration is not uniform.

The equations of motion

For motion with constant acceleration in a straight line, WJEC uses four equations (often called the $suvat$ equations):

Here $u$ is the initial velocity, $v$ the final velocity, $a$ the acceleration, $x$ the displacement and $t$ the time. Choose the equation that contains the three known quantities and the one unknown you want, so you avoid solving simultaneous equations.

Projectile motion

A projectile moves under gravity alone. Resolve the launch velocity into horizontal and vertical components using $u_x = u\cos\theta$ and $u_y = u\sin\theta$. The horizontal velocity is constant because no horizontal force acts; the vertical motion has constant acceleration $g \approx 9.81\,\text{m s}^{-2}$ downward. Time is the bridge: solve the vertical motion for the time of flight, then feed that time into the horizontal equation for the range.

Examples in context

Example 1. A long jumper. An athlete leaves the board at $9.0\,\text{m s}^{-1}$ at $20^{\circ}$ above the horizontal. The vertical component is $u_y = 9.0\sin 20^{\circ} = 3.08\,\text{m s}^{-1}$, so the time to the peak is $t = u_y/g = 0.31\,\text{s}$ and the full flight is $0.63\,\text{s}$. The horizontal component is $u_x = 9.0\cos 20^{\circ} = 8.46\,\text{m s}^{-1}$, giving a range of $8.46 \times 0.63 = 5.3\,\text{m}$. This shows why sprint speed matters more than jump angle for human jumpers.

Example 2. A dropped sensor package. A research package released from a hovering drone at $80\,\text{m}$ falls with $u = 0$. Using $y = \tfrac{1}{2}gt^2$, $t = \sqrt{2 \times 80 / 9.81} = 4.04\,\text{s}$, and it lands at $v = gt = 39.6\,\text{m s}^{-1}$. Adding any horizontal drift simply rides on top of this vertical answer because the two directions are independent.

Try this

Q1. A car accelerates from rest at $2.0\,\text{m s}^{-2}$ for $5.0\,\text{s}$. Find its final velocity and the distance travelled. [3 marks]

• Cue. $v = 0 + 2.0\times5.0 = 10\,\text{m s}^{-1}$; $x = \frac{1}{2}\times2.0\times5.0^2 = 25\,\text{m}$.

Q2. Explain why the horizontal velocity of a projectile stays constant. [2 marks]

• Cue. No horizontal force acts (neglecting air resistance), so there is no horizontal acceleration.

Q3. A ball is thrown vertically upward at $20\,\text{m s}^{-1}$. Find the maximum height reached. [3 marks]

• Cue. $v^2 = u^2 + 2ax$ with $v = 0$: $0 = 20^2 - 2(9.81)x$, so $x = 400/19.62 = 20.4\,\text{m}$.