Kinematics of motion in a straight line: displacement, velocity, acceleration, equations of motion

🚀 Mechanics (M1): Kinematics of Motion in a Straight Line

📍 Displacement

Displacement, denoted by $\Delta \mathbf{s}$, is the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. Analogy: Think of a runner on a straight track. If they start at the 0 m mark and finish at the 30 m mark, their displacement is +30 m in the direction of the track. If they finish at the 10 m mark, the displacement is +10 m. If they finish at the -5 m mark (behind the start), the displacement is –5 m.

📈 Velocity

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

$$\mathbf{v} = \frac{d\mathbf{s}}{dt}$$

If an object moves at a constant speed of 5 m s⁻¹ to the right, its velocity is $\mathbf{v}=+5\,\text{m\,s}^{-1}$. Example: A cyclist travels 20 km east in 2 h. $$\mathbf{v} = \frac{20\,\text{km}}{2\,\text{h}} = 10\,\text{km\,h}^{-1}\text{ east}$$

⚡ Acceleration

Acceleration is the rate of change of velocity. It is also a vector.

$$\mathbf{a} = \frac{d\mathbf{v}}{dt}$$

If a car speeds up from 0 to 20 m s⁻¹ in 5 s, its average acceleration is $$\mathbf{a}_{\text{avg}} = \frac{20\,\text{m\,s}^{-1} - 0}{5\,\text{s}} = 4\,\text{m\,s}^{-2}$$

📐 Equations of Motion (Constant Acceleration)

When acceleration is constant, the following equations hold (often called the “SUVAT” equations).

1. $\displaystyle \mathbf{v} = \mathbf{u} + \mathbf{a}t$ – Final velocity after time $t$.
2. $\displaystyle \mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$ – Displacement after time $t$.
3. $\displaystyle \mathbf{v}^2 = \mathbf{u}^2 + 2\mathbf{a}\mathbf{s}$ – Relates velocities and displacement without time.

Where:

• $\mathbf{u}$ – Initial velocity
• $\mathbf{v}$ – Final velocity
• $\mathbf{a}$ – Acceleration
• $t$ – Time
• $\mathbf{s}$ – Displacement

Equation	Use
$\,\mathbf{v} = \mathbf{u} + \mathbf{a}t$	Find final velocity when time and acceleration are known.
$\,\mathbf{s} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$	Find displacement when time, initial velocity and acceleration are known.
$\,\mathbf{v}^2 = \mathbf{u}^2 + 2\mathbf{a}\mathbf{s}$	Find final velocity or displacement when time is not needed.

🏃‍♂️ Example Problem

A skateboarder starts from rest and accelerates at $2\,\text{m\,s}^{-2}$ for $10\,\text{s}$.

1. Find the final velocity.
2. Find the displacement.

1. $\displaystyle \mathbf{v} = 0 + (2\,\text{m\,s}^{-2})(10\,\text{s}) = 20\,\text{m\,s}^{-1}$ 2. $\displaystyle \mathbf{s} = 0 \times 10\,\text{s} + \tfrac{1}{2}(2\,\text{m\,s}^{-2})(10\,\text{s})^2 = 100\,\text{m}$

So the skateboarder reaches 20 m s⁻¹ and travels 100 m in 10 s.

💡 Key Takeaways

• Displacement, velocity, and acceleration are all vectors.
• Use the correct sign (positive or negative) to indicate direction.
• For constant acceleration, the SUVAT equations give a quick way to solve problems.
• Always check units – they must be consistent across all quantities.