Kinematics of motion in 2 dimensions: displacement, velocity, acceleration, projectile motion

Mechanics (M2) – Kinematics in 2D 🚀

1️⃣ Displacement

Displacement is the straight‑line vector from the start point to the end point. It has both magnitude and direction.

Analogy: Imagine walking from your home to a friend's house. The distance you actually walk (path length) is different from the straight‑line distance between the two houses.

Formula: $\displaystyle \mathbf{s}= \mathbf{r}_f-\mathbf{r}_i$

  • Vector notation: $\mathbf{s} = (x_f-x_i,\, y_f-y_i)$
  • Magnitude: $|\mathbf{s}| = \sqrt{(x_f-x_i)^2+(y_f-y_i)^2}$
Exam Tip: Always keep track of the direction (use unit vectors or angles) when you write displacement components.

2️⃣ Velocity

Velocity is the rate of change of displacement. It is a vector.

Analogy: Think of a cyclist who pedals faster – their velocity increases.

Formula (instantaneous): $\displaystyle \mathbf{v} = \frac{d\mathbf{s}}{dt}$

For constant velocity: $\displaystyle \mathbf{v} = \frac{\mathbf{s}}{t}$

  1. Component form: $v_x = \frac{dx}{dt}$, $v_y = \frac{dy}{dt}$
  2. Magnitude: $|\mathbf{v}| = \sqrt{v_x^2+v_y^2}$
Exam Tip: When solving for velocity, check if the motion is uniform (constant speed) or non‑uniform (changing speed).

3️⃣ Acceleration

Acceleration is the rate of change of velocity.

Analogy: If you press the gas pedal harder, the car’s acceleration increases.

Formula (instantaneous): $\displaystyle \mathbf{a} = \frac{d\mathbf{v}}{dt}$

For constant acceleration: $\displaystyle \mathbf{a} = \frac{\Delta\mathbf{v}}{\Delta t}$

  • Component form: $a_x = \frac{dv_x}{dt}$, $a_y = \frac{dv_y}{dt}$
  • Magnitude: $|\mathbf{a}| = \sqrt{a_x^2+a_y^2}$
Exam Tip: Remember that in projectile motion, only the vertical component of acceleration is non‑zero (gravity).

4️⃣ Projectile Motion 🎯

A projectile follows a parabolic path under the influence of gravity alone (ignoring air resistance).

Key assumptions:

  • Initial position at the origin: $\mathbf{r}_0 = (0,0)$
  • Initial velocity: $\mathbf{v}_0 = (v_{0x}, v_{0y})$
  • Acceleration: $\mathbf{a} = (0, -g)$, where $g \approx 9.81\,\text{m/s}^2$

Equations of motion:

  • Horizontal: $x = v_{0x}\,t$
  • Vertical: $y = v_{0y}\,t - \tfrac{1}{2}gt^2$
  • Velocity components: $v_x = v_{0x}$, $v_y = v_{0y} - g t$

Launch angle $\theta$ gives: $v_{0x} = v_0\cos\theta$, $v_{0y} = v_0\sin\theta$.

Parameter Symbol Formula Example
Time of flight $T$ $T = \frac{2v_0\sin\theta}{g}$ $v_0=20\,\text{m/s},\;\theta=45^\circ \Rightarrow T \approx 2.88\,\text{s}$
Maximum height $H$ $H = \frac{v_0^2\sin^2\theta}{2g}$ $v_0=20\,\text{m/s},\;\theta=45^\circ \Rightarrow H \approx 20.4\,\text{m}$
Range $R$ $R = \frac{v_0^2\sin 2\theta}{g}$ $v_0=20\,\text{m/s},\;\theta=45^\circ \Rightarrow R \approx 40.8\,\text{m}$
Exam Tip: When a question asks for the maximum height or range, always start from the kinematic equations and eliminate time where possible.

Key Take‑aways 📚

  1. Displacement is a vector; use components.
  2. Velocity and acceleration are derivatives of displacement and velocity respectively.
  3. In projectile motion, horizontal motion is uniform; vertical motion is uniformly accelerated.
  4. Always check units and keep track of signs (positive upwards, negative downwards).
Final Exam Tip: Practice translating word problems into equations, then solve step by step. Remember to show all work – the examiner can award partial marks for correct reasoning even if the final number is slightly off.

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