Equations of motion

1. Overview

Kinematics is the study of motion without considering the forces that cause it. In the Cambridge 9702 syllabus, the core principle is that motion is described by the precise mathematical relationship between displacement, velocity, and acceleration over time. A fundamental requirement for success in this topic is the rigorous application of vector versus scalar distinctions. While distance and speed are scalars, displacement, velocity, and acceleration are vectors; therefore, the direction of motion—represented by algebraic signs (+ or –)—is as critical as the magnitude. Most analytical problems in this section involve uniformly accelerated motion (constant acceleration), which permits the use of the kinematic equations (SUVAT). However, students must also be prepared to describe non-uniform motion qualitatively, particularly in the context of air resistance and terminal velocity.

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Key Definitions

To earn full marks in "Define" questions, you must include the specific keywords underlined in Cambridge mark schemes.

• Distance: The total path length travelled by an object. (Scalar; SI unit: $\text{m}$)
• Displacement ($s$): The distance moved in a specified direction from a fixed reference point. It is the straight-line separation between two points. (Vector; SI unit: $\text{m}$)
• Speed: The distance travelled per unit time. (Scalar; SI unit: $\text{m,s}^{-1}$)
• Velocity ($v$): The rate of change of displacement. (Vector; SI unit: $\text{m,s}^{-1}$)
  • Note: Mathematically, $v = \frac{\Delta s}{\Delta t}$.
• Acceleration ($a$): The rate of change of velocity. (Vector; SI unit: $\text{m,s}^{-2}$)
  • Note: Mathematically, $a = \frac{\Delta v}{\Delta t}$. Acceleration occurs if either the magnitude (speed) or the direction of velocity changes.

Average vs. Instantaneous Quantities:

• Average Velocity: The total displacement divided by the total time taken. It does not account for fluctuations during the journey.
• Instantaneous Velocity: The velocity of an object at a specific moment in time. On a displacement-time graph, this is the gradient of the tangent to the curve at that specific time.

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Content

3.1 Graphical Representation of Motion

Graphs are a primary tool for analyzing motion. You must be able to interpret the shape, calculate the gradient, and determine the area under the curve.

Displacement–Time ($s-t$) Graphs

• Gradient: Represents the velocity.
• Horizontal Line: The object is stationary (velocity = 0).
• Straight Diagonal Line: The object is moving with constant velocity.
• Curve (Increasing Gradient): The object is accelerating.
• Curve (Decreasing Gradient): The object is decelerating.
• Negative Gradient: The object is moving back toward the reference point (negative velocity).

Velocity–Time ($v-t$) Graphs

• Gradient: Represents the acceleration.
  • A straight line indicates uniform acceleration.
  • A curve indicates non-uniform acceleration.
• Area Under the Graph: Represents the displacement.
  • To find the total distance, add the absolute values of the areas (treat negative areas as positive).
  • To find the displacement, subtract the area below the time axis from the area above it.
• Horizontal Line: The object is moving with constant velocity (acceleration = 0).
• Intercepts: The y-intercept is the initial velocity ($u$); the x-intercept is the time when the object is instantaneously at rest.

Acceleration–Time ($a-t$) Graphs

• Area Under the Graph: Represents the change in velocity ($\Delta v = v - u$).
• Horizontal Line (not on x-axis): The object has constant acceleration.
• Line on the x-axis ($a=0$): The object is moving with constant velocity.

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3.2 Deriving Equations of Motion (SUVAT)

You are required to derive these equations from the definitions of velocity and acceleration. These equations only apply when acceleration is constant.

Derivation 1: From the definition of acceleration Acceleration is the rate of change of velocity: $$a = \frac{v - u}{t}$$ Rearranging for $v$: $$v = u + at$$

Derivation 2: From the average velocity For constant acceleration, the average velocity is the arithmetic mean of the initial and final velocities: $$\text{Average velocity} = \frac{u + v}{2}$$ Since displacement $s = \text{average velocity} \times t$: $$s = \frac{(u + v)}{2}t$$

Derivation 3: Substituting (1) into (2) To eliminate $v$, substitute $v = u + at$ into equation (2): $$s = \frac{u + (u + at)}{2}t$$ $$s = \frac{2u + at}{2}t$$ $$s = ut + \frac{1}{2}at^2$$

Derivation 4: Eliminating $t$ Rearrange $v = u + at$ to get $t = \frac{v - u}{a}$. Substitute this into equation (2): $$s = \left(\frac{v + u}{2}\right) \left(\frac{v - u}{a}\right)$$ $$s = \frac{v^2 - u^2}{2a}$$ $$v^2 = u^2 + 2as$$

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3.3 Free Fall and the Acceleration of Free Fall ($g$)

An object is in "free fall" when the only force acting on it is gravity (weight).

• In the absence of air resistance, all objects near the Earth's surface fall with a constant acceleration $g = 9.81 , \text{m,s}^{-2}$.
• Sign Convention: You must be consistent.
  • If you choose Up as positive (+): $a = -9.81 , \text{m,s}^{-2}$.
  • If you choose Down as positive (+): $a = +9.81 , \text{m,s}^{-2}$.

Worked Example 1 — Vertical Projection

A stone is thrown vertically upwards from the top of a cliff with an initial velocity of $15.0 , \text{m,s}^{-1}$. The cliff is $20.0 , \text{m}$ high. Calculate the velocity of the stone just before it hits the ground at the base of the cliff.

1. Define Sign Convention: Let upwards be positive (+).
2. Identify Variables:
   • $u = +15.0 , \text{m,s}^{-1}$
   • $s = -20.0 , \text{m}$ (The stone ends up 20m below the starting point)
   • $a = -9.81 , \text{m,s}^{-2}$
   • $v = ?$
3. Select Equation: $v^2 = u^2 + 2as$
4. Substitution: $$v^2 = (15.0)^2 + 2(-9.81)(-20.0)$$ $$v^2 = 225 + 392.4 = 617.4$$
5. Solve: $$v = \sqrt{617.4} = \pm 24.847...$$
6. Final Answer: Since the stone is moving downwards, $v = -24.8 , \text{m,s}^{-1}$ (or $24.8 , \text{m,s}^{-1}$ downwards).

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3.4 Determining $g$ Experimentally

The standard A-Level experiment uses a falling object and an electronic timer to minimize human reaction time errors.

Method: Electromagnet and Trapdoor

1. Apparatus: Electromagnet, steel ball bearing, trapdoor with switch, electronic timer, meter rule.
2. Procedure:
   • Measure the height $h$ from the bottom of the ball to the trapdoor using a meter rule.
   • Release the ball by switching off the electromagnet. This action starts the timer.
   • The ball hits the trapdoor, breaking the circuit and stopping the timer. Record time $t$.
   • Repeat for at least six different heights $h$.
3. Analysis:
   • From $s = ut + \frac{1}{2}at^2$, with $u = 0$ and $s = h$:
   • $h = \frac{1}{2}gt^2$
   • Plot a graph of $h$ (y-axis) against $t^2$ (x-axis).
   • The gradient of the line of best fit is $\frac{g}{2}$.
   • Calculate $g = 2 \times \text{gradient}$.
4. Evaluation:
   • Systematic Error: Residual magnetism in the electromagnet may cause a delay in release, making $t$ recorded larger than the actual fall time. This results in an underestimate of $g$.
   • Random Error: Parallax error when reading the meter rule.
   • Improvement: Use a small, dense ball to minimize the effect of air resistance.

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3.5 Projectile Motion

Projectile motion occurs when an object has a uniform velocity in one direction (usually horizontal) and a uniform acceleration in a perpendicular direction (usually vertical gravity).

The Principle of Independence: The horizontal and vertical components of motion do not affect each other. They are linked only by time ($t$).

Strategy for Solving Projectile Problems:

1. Resolve the initial velocity $u$ into components:
   • $u_x = u \cos \theta$
   • $u_y = u \sin \theta$
2. Horizontal Motion ($a_x = 0$):
   • Velocity is constant: $v_x = u_x$.
   • Displacement: $x = u_x t$.
3. Vertical Motion ($a_y = -9.81 , \text{m,s}^{-2}$):
   • Use SUVAT equations: $v_y = u_y + at$ and $s_y = u_y t + \frac{1}{2}at^2$.

Worked Example 2 — Horizontal Projection

A rescue plane flying horizontally at $60.0 , \text{m,s}^{-1}$ at a height of $200 , \text{m}$ drops a relief package. Calculate the horizontal distance the package travels before hitting the ground.

1. Vertical Analysis (to find time $t$):
   • $u_y = 0$ (Initial motion is purely horizontal)
   • $s_y = -200 , \text{m}$ (Taking up as positive)
   • $a_y = -9.81 , \text{m,s}^{-2}$
   • $s_y = u_y t + \frac{1}{2}a_y t^2 \implies -200 = 0 + \frac{1}{2}(-9.81)t^2$
   • $t^2 = \frac{-200}{-4.905} = 40.77...$
   • $t = 6.385 , \text{s}$
2. Horizontal Analysis:
   • $u_x = 60.0 , \text{m,s}^{-1}$
   • $x = u_x \times t$
   • $x = 60.0 \times 6.385 = 383.1...$
3. Final Answer: $x = 383 , \text{m}$ (to 3 s.f.)

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Key Equations

Equation	Description	Data Sheet?
$v = u + at$	Final velocity	Yes
$s = ut + \frac{1}{2}at^2$	Displacement (given $u$)	Yes
$s = vt - \frac{1}{2}at^2$	Displacement (given $v$)	No
$s = \frac{(u+v)}{2}t$	Displacement (average velocity)	Yes
$v^2 = u^2 + 2as$	Velocity-displacement	Yes
$v_x = u \cos \theta$	Horizontal velocity component	No
$v_y = u \sin \theta$	Vertical velocity component	No

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Common Mistakes to Avoid

• ❌ Wrong: Using $g = 10 , \text{m,s}^{-2}$ or $g = 9.8 , \text{m,s}^{-2}$. ✓ Right: Always use $g = 9.81 , \text{m,s}^{-2}$ as specified in the 9702 Data Booklet.
• ❌ Wrong: Mixing horizontal and vertical components in the same SUVAT equation. ✓ Right: Keep $x$ and $y$ variables strictly separate. Only $t$ is interchangeable.
• ❌ Wrong: Assuming $v=0$ at the end of a fall when calculating impact velocity. ✓ Right: $v$ is the velocity the instant before impact. After impact, the acceleration is no longer $g$.
• ❌ Wrong: Forgetting that displacement is a vector. ✓ Right: If an object is thrown up and falls past its starting point, $s$ must be negative.
• ❌ Wrong: Using SUVAT for motion with significant air resistance. ✓ Right: SUVAT only applies if acceleration is constant. Air resistance causes acceleration to decrease as speed increases.

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Exam Tips

1. The SUVAT List: Before starting any calculation, write $s, u, v, a, t$ in the margin and fill in the knowns. This prevents "formula hunting" and identifies the missing variable.
2. State the Direction: Always start a projectile or vertical motion problem by writing "Taking upwards as positive". This clarifies your sign convention for the examiner.
3. Significant Figures: Final answers should usually be given to 3 significant figures. However, keep the full value in your calculator for intermediate steps to avoid rounding errors.
4. Bouncing Ball Graphs: In $v-t$ graphs for a bouncing ball, the gradient (acceleration) remains constant at $-9.81 , \text{m,s}^{-2}$ (straight parallel lines), but the velocity "jumps" from negative to positive at the moment of impact.
5. Peak of Flight: At the maximum height of a projectile, the vertical velocity is zero ($v_y = 0$), but the horizontal velocity is NOT zero ($v_x = u \cos \theta$). The acceleration at the peak is still $9.81 , \text{m,s}^{-2}$ downwards.