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

Cambridge AS & A‑Level Mathematics 9709 – Mechanics (M1)

4.1 Forces & Equilibrium

Key Definitions

• Force ( \(\mathbf{F}\) ): A vector quantity capable of changing the state of motion of a particle.
• Resultant force: Vector sum of all forces acting on a particle.
• Equilibrium condition: \(\displaystyle\sum\mathbf{F}= \mathbf{0}\) (no net force ⇒ no acceleration).

Components of a Force

Resolve any force into orthogonal components using trigonometry. For a force \(\mathbf{F}\) making an angle \(\theta\) with the positive \(x\)‑axis:

Limiting Friction & Limiting Equilibrium

• Limiting friction (\(F_{\text{lim}}\)): Maximum static friction before motion starts, \(F_{\text{lim}}=\mu_s N\) where \(\mu_s\) is the coefficient of static friction and \(N\) the normal reaction.
• Limiting equilibrium: The state just before motion begins; the resultant force is zero but the frictional force is at its limiting value.
• For a rough horizontal surface: \(\displaystyle\sum F_x = 0 \;\Rightarrow\; T - F_{\text{lim}} = 0\) (example of a pulling force \(T\) on a block).

Smooth vs Rough Contact

Worked Example – Hanging Weight (Equilibrium)

A mass \(m=2.0\;\text{kg}\) is suspended from a light string. The only forces are the weight \(\mathbf{W}=mg\) (downward) and the tension \(\mathbf{T}\) (upward).

• Equilibrium: \(\displaystyle T - mg = 0\)
• \(T = mg = 2.0\times9.8 = 19.6\;\text{N}\) upward.

4.2 Kinematics of Motion in a Straight Line (Core topic)

Learning Objectives

1. Define displacement, velocity and acceleration for one‑dimensional motion.
2. Distinguish between average and instantaneous quantities.
3. Apply the three constant‑acceleration equations.
4. Interpret and construct displacement‑time, velocity‑time and acceleration‑time graphs.
5. Derive the equations of motion using calculus (for constant \(a\)).
6. Adopt a consistent sign convention and avoid common sign errors.

Key Concepts

• Straight‑line motion: Motion confined to a single axis (taken as the \(x\)-axis).
• Scalar vs vector: Displacement \(\mathbf{s}\), velocity \(\mathbf{v}\) and acceleration \(\mathbf{a}\) are vectors; speed is the magnitude of velocity.
• Constant acceleration: When \(a\) does not vary with time, the standard kinematic equations apply.
• Sign convention: Choose a positive direction (e.g. rightward or upward). Quantities opposite to this direction are negative.

Definitions

Displacement \(s\) (vector) – change in position: \(s = x_{\text{f}}-x_{\text{i}}\).
Average velocity \(\bar v\) over \(\Delta t\): \(\displaystyle\bar v=\frac{\Delta s}{\Delta t}\).
Instantaneous velocity \(v\): \(\displaystyle v=\frac{dx}{dt}\).
Average acceleration \(\bar a\) over \(\Delta t\): \(\displaystyle\bar a=\frac{\Delta v}{\Delta t}\).
Instantaneous acceleration \(a\): \(\displaystyle a=\frac{dv}{dt}= \frac{d^{2}x}{dt^{2}}\).

Link to Calculus

From the definition of velocity and acceleration:

For constant acceleration (\(a\) = constant):

1. Integrate \(a = dv/dt\) → \(\displaystyle v = u + at\) (where \(u\) is the velocity at \(t=0\)).
2. Integrate \(v = dx/dt\) using the result above → \(\displaystyle s = ut + \tfrac12 at^{2}\) (where \(s\) is the displacement during time \(t\)).
3. Eliminate \(t\) between the two equations to obtain the third relation: \(\displaystyle v^{2}=u^{2}+2as\).

Equations of Motion (Constant Acceleration)

Graphical Interpretation (Constant \(a\))

• Velocity–time graph: Straight line; slope = \(a\), \(y\)-intercept = \(u\).
• Displacement–time graph: Parabolic curve; curvature proportional to \(|a|\).
• Acceleration–time graph: Horizontal line at the constant value \(a\) (zero slope).
• The area under a velocity–time graph between two times equals the signed displacement in that interval.
• The area under an acceleration–time graph gives the change in velocity.

Worked Example – Uniformly Accelerating Car

Problem: A car starts from rest and accelerates uniformly at \(2.5\;\text{m s}^{-2}\) for \(8\;\text{s}\). Find (i) the final speed, (ii) the distance travelled, and (iii) the speed after \(5\;\text{s}\).

1. Given: \(u=0\;\text{m s}^{-1},\; a=2.5\;\text{m s}^{-2},\; t=8\;\text{s}\).
2. Final speed: \(\displaystyle v = u + at = 0 + (2.5)(8) = 20\;\text{m s}^{-1}\).
3. Distance: \(\displaystyle s = ut + \tfrac12 at^{2}=0 + \tfrac12(2.5)(8^{2}) = 0.5\times2.5\times64 = 80\;\text{m}\).
4. Speed after \(5\;\text{s}\): \(\displaystyle v_{5}=u + a(5)=0+(2.5)(5)=12.5\;\text{m s}^{-1}\).

Answers: (i) \(20\;\text{m s}^{-1}\) (ii) \(80\;\text{m}\) (iii) \(12.5\;\text{m s}^{-1}\).

Common Pitfalls

• Confusing displacement (signed vector) with distance (scalar, always positive).
• Using the wrong sign for acceleration when the object is decelerating; remember the sign follows the chosen positive direction.
• Applying the three equations when acceleration is not constant – in such cases calculus, numerical methods, or piece‑wise analysis are required.
• Neglecting the area‑under‑curve interpretation for displacement and velocity.

Practice Questions

1. A particle moves along the \(x\)-axis with constant acceleration \(a=-3\;\text{m s}^{-2}\). Its initial velocity is \(u=12\;\text{m s}^{-1}\).
   • (a) Time taken to come to rest.
   • (b) Distance covered in that time.
2. A ball is thrown vertically upward with an initial speed of \(15\;\text{m s}^{-1}\). Using \(g=9.8\;\text{m s}^{-2}\) (downward), determine
   • (a) The maximum height reached.
   • (b) The total time the ball remains in the air.
3. Sketch the velocity–time graph for a car that
   • (i) starts from rest and accelerates uniformly for \(4\;\text{s}\),
   • (ii) then travels at constant speed for \(6\;\text{s}\),
   • (iii) finally decelerates uniformly to rest in \(3\;\text{s}\).
4. Using the sign convention “upward is positive”, a particle is projected upward with \(u=20\;\text{m s}^{-1}\). Find the velocity after \(3\;\text{s}\) and the displacement during that interval.

4.3 Momentum

Definition

Linear momentum of a particle of mass \(m\) moving with velocity \(\mathbf{v}\) is \(\displaystyle\mathbf{p}=m\mathbf{v}\). Momentum is a vector and is conserved in the absence of external forces.

One‑Dimensional Collisions

• Elastic collision: Both momentum and kinetic energy are conserved.
• Inelastic collision: Momentum is conserved, kinetic energy is not. In a perfectly inelastic collision the bodies stick together.

Worked Example – Perfectly Inelastic Collision

Two carts on a frictionless track: Cart A: \(m_A=1.5\;\text{kg}\), \(v_A=4\;\text{m s}^{-1}\) (rightward). Cart B: \(m_B=2.0\;\text{kg}\), initially at rest. After they stick together, find the common speed \(v\).

4.4 Newton’s Laws of Motion

Application – Inclined Plane (Smooth)

Block of mass \(m=3\;\text{kg}\) on a smooth plane inclined at \(\theta=30^{\circ}\). Choose the direction down the plane as positive.

• Component of weight along the plane: \(\displaystyle mg\sin\theta = 3\times9.8\times\sin30^{\circ}=14.7\;\text{N}\).
• Net force \(F_{\text{net}} = mg\sin\theta\) (no friction).
• Acceleration: \(\displaystyle a = \frac{F_{\text{net}}}{m}= \frac{14.7}{3}=4.9\;\text{m s}^{-2}\) down the plane.

Typical Problems

• Resolving forces on an object on a rough incline (include limiting friction).
• Finding tension in a string supporting a mass with acceleration.
• Analyzing systems of particles using \(\sum\mathbf{F}=m\mathbf{a}\) for each particle and \(\sum\mathbf{F}_{\text{ext}} = M\mathbf{a}_{\text{CM}}\) for the whole system.

4.5 Energy, Work & Power

Work Done by a Constant Force

If a constant force \(\mathbf{F}\) acts through a displacement \(\mathbf{s}\) in the direction of the force, the work done is

Work‑Energy Theorem

The net work done on a particle equals the change in its kinetic energy:

Power

Power is the rate at which work is done:

Link to Kinematics

For a body of mass \(m\) under a constant net force \(\mathbf{F}=m\mathbf{a}\):

• Integrating \(\mathbf{F}=m\mathbf{a}\) gives the constant‑acceleration equations of motion (Section 4.2).
• The work‑energy theorem provides an alternative route to the equation \(v^{2}=u^{2}+2as\) by setting \(\displaystyle F s = \tfrac12 m(v^{2}-u^{2})\).

Summary

Mastering straight‑line kinematics equips you with the tools required for the remainder of the Mechanics (M1) syllabus. You should now be able to:

• Define and use displacement, velocity and acceleration, both average and instantaneous.
• Apply the three constant‑acceleration equations correctly, recognising the underlying calculus.
• Read, construct and interpret displacement‑time, velocity‑time and acceleration‑time graphs, using the area‑under‑curve method.
• Adopt a clear sign convention and avoid sign‑related mistakes, especially when dealing with deceleration or motion opposite to the chosen positive direction.
• Analyse forces in equilibrium (including limiting friction), use Newton’s laws for dynamics, and employ momentum and energy principles where appropriate.

These foundations will be essential for later topics such as projectile motion, circular motion, and the more advanced dynamics covered in Papers 4 and 5.