Kinematics: Motion in One and Two Dimensions

Canada Grade 11 Physics Revision Notes

These revision notes are designed for Canadian Grade 11 Physics students, with a focus on understanding motion, vectors, graphs, acceleration, kinematic equations, and projectile motion.

1. What Is Kinematics?

Kinematics is the branch of physics that describes the motion of objects without focusing on the forces that cause that motion. In Grade 11 Physics, kinematics provides the foundation for understanding more advanced topics such as forces, energy, momentum, and circular motion.

When studying kinematics, we ask questions such as:

  • Where is an object?
  • How far has it travelled?
  • How quickly is it moving?
  • In which direction is it moving?
  • Is its velocity changing?
  • How long does the motion take?

Kinematics allows us to answer these questions using measurements, graphs, vectors, and mathematical equations.

2. Position, Distance and Displacement

Position

Position describes where an object is located relative to a chosen reference point.

For example, if a student is 20 m east of the school entrance, the school entrance is the reference point and the student's position is 20 m east.

Position is a vector quantity because it includes both magnitude and direction.

Distance

Distance is the total length of the path travelled by an object. Distance is a scalar quantity, so it has magnitude but no direction.

Suppose a student walks 30 m east and then 20 m west. The total distance is:

\[ d = 30 + 20 = 50\,\text{m} \]

Therefore, the student travelled 50 m.

Displacement

Displacement is the change in position from the starting point to the final point.

For the same student:

\[ \Delta x = 30 - 20 = 10\,\text{m} \]

Therefore, the displacement is 10 m east.

Quantity Meaning Type
Distance Total path travelled Scalar
Displacement Change in position Vector
Exam Tip: Distance can never be negative. Displacement can be positive, negative, or zero depending on the coordinate system.

If you walk around a track and return to your starting point, your distance is greater than zero, but your displacement is zero.

3. Scalars and Vectors

A scalar quantity has magnitude only.

Examples include:

  • Distance
  • Speed
  • Time
  • Mass
  • Temperature

A vector quantity has both magnitude and direction.

Examples include:

  • Displacement
  • Velocity
  • Acceleration
  • Force

Consider the difference between:

80 km/h — speed

80 km/h east — velocity

The direction makes velocity a vector quantity.

4. Choosing a Coordinate System

Before solving a kinematics problem, choose a positive direction. This makes signs in your equations much easier to manage.

For example, if east is positive:

  • East = positive
  • West = negative

Similarly, for vertical motion you might choose:

  • Up = positive
  • Down = negative

If an object moves 15 m east:

\[ \Delta x = +15\,\text{m} \]

If it moves 15 m west:

\[ \Delta x = -15\,\text{m} \]
Rahul Sir's Rule: Never start a kinematics calculation before deciding your positive direction.

5. Speed and Velocity

Speed

Speed describes how quickly distance is covered.

\[ v = \frac{d}{t} \]

where:

  • v = speed
  • d = distance
  • t = time

The SI unit of speed is:

\[ \text{m/s} \]

Speed is a scalar quantity.

Velocity

Velocity describes the rate at which displacement changes.

\[ v = \frac{\Delta x}{\Delta t} \]

Velocity is a vector quantity because it includes direction.

Example: A car travels 100 m east in 5 seconds.

\[ v = \frac{100}{5} \]
\[ v = 20\,\text{m/s} \]

Therefore, the velocity is 20 m/s east.

6. Average Speed and Average Velocity

Average Speed

\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]

Average Velocity

\[ v_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} \]

Average speed and average velocity are not necessarily the same.

Imagine a runner completing one full lap and returning to the starting point. The runner has travelled a considerable distance, but the final position is the same as the initial position.

Therefore:

\[ \text{Displacement} = 0 \]
\[ v_{\text{avg}} = 0 \]

However, the average speed is greater than zero.

7. Acceleration

Acceleration is the rate at which velocity changes.

\[ a = \frac{\Delta v}{\Delta t} \]

or:

\[ a = \frac{v_f-v_i}{t} \]

where:

  • a = acceleration
  • vi = initial velocity
  • vf = final velocity
  • t = time

The SI unit of acceleration is:

\[ \text{m/s}^2 \]

Acceleration can occur when:

  1. Speed increases.
  2. Speed decreases.
  3. Direction changes.
  4. Both speed and direction change.
Important: An object can have constant speed and still have acceleration if its direction changes.

8. Positive and Negative Acceleration

Negative acceleration does not automatically mean that an object is slowing down.

The meaning depends on the chosen positive direction and the object's velocity.

Suppose east is positive and a car travels east:

\[ v = +20\,\text{m/s} \]

If:

\[ a = -5\,\text{m/s}^2 \]

the velocity becomes less positive, so the car slows down.

However, if the car is travelling west:

\[ v = -20\,\text{m/s} \]

and:

\[ a = -5\,\text{m/s}^2 \]

velocity and acceleration point in the same direction, so the object speeds up.

Remember:
Velocity and acceleration in the same direction → speed increases.
Velocity and acceleration in opposite directions → speed decreases.

9. Uniform and Non-Uniform Motion

Uniform Motion

An object has uniform motion when its velocity remains constant. Therefore:

\[ a = 0 \]

For example, a car travelling east at a constant 15 m/s has constant velocity.

Uniform Acceleration

An object has uniform acceleration when its acceleration remains constant.

For example:

\[ a = 2\,\text{m/s}^2 \]

This means the velocity changes by 2 m/s every second.

Time Velocity
0 s 5 m/s
1 s 7 m/s
2 s 9 m/s
3 s 11 m/s

10. The Four Kinematic Equations

The four major kinematic equations are used for motion involving constant acceleration.

Equation 1

\[ v_f = v_i + at \]

This equation is useful when displacement is not required.

Equation 2

\[ \Delta x = v_i t + \frac{1}{2}at^2 \]

This equation is useful when final velocity is not required.

Equation 3

\[ v_f^2 = v_i^2 + 2a\Delta x \]

This equation is particularly useful when time is not given.

Equation 4

\[ \Delta x = \frac{v_i+v_f}{2}t \]

This equation is useful when acceleration is not directly required.

Exam Strategy: Before choosing an equation, identify which quantities are known and which quantity you need to find.

11. Worked Example: Constant Acceleration

A car starts from rest and accelerates at 3 m/s² for 5 seconds. Find its final velocity.

Step 1: Identify the known values

\[ v_i = 0 \]
\[ a = 3\,\text{m/s}^2 \]
\[ t = 5\,\text{s} \]

Step 2: Choose the equation

\[ v_f = v_i + at \]

Step 3: Substitute

\[ v_f = 0 + (3)(5) \]
\[ \boxed{v_f = 15\,\text{m/s}} \]

The final velocity is 15 m/s in the direction of the acceleration.

12. Motion Graphs

Graphs are an essential part of kinematics. The three major graphs are:

  • Position-time graph
  • Velocity-time graph
  • Acceleration-time graph

Position-Time Graph

The slope of a position-time graph represents velocity.

\[ v = \frac{\Delta x}{\Delta t} \]

Interpretation:

  • Positive slope → positive velocity
  • Negative slope → negative velocity
  • Zero slope → object is at rest
  • Steeper slope → greater speed

Velocity-Time Graph

The slope of a velocity-time graph represents acceleration.

\[ a = \frac{\Delta v}{\Delta t} \]

The area under a velocity-time graph represents displacement.

Acceleration-Time Graph

The area under an acceleration-time graph represents the change in velocity.

\[ \Delta v = a\Delta t \]

Therefore:

\[ \Delta v = v_f-v_i \]
Graph Memory Trick:
Position → slope gives velocity.
Velocity → slope gives acceleration.
Velocity → area gives displacement.
Acceleration → area gives change in velocity.

13. Motion in Two Dimensions

One-dimensional motion occurs along a single straight line.

Examples include:

  • A car travelling along a straight road
  • An elevator moving vertically
  • An object falling vertically

Two-dimensional motion occurs when an object moves in two perpendicular directions.

Examples include:

  • A ball thrown through the air
  • A person walking northeast
  • A plane travelling with wind
  • A projectile launched at an angle

For two-dimensional problems, motion is normally separated into horizontal and vertical components.

14. Vectors in Two Dimensions

A vector can be separated into horizontal and vertical components. Suppose a vector with magnitude V makes an angle θ with the horizontal.

The horizontal component is:

\[ V_x = V\cos\theta \]

The vertical component is:

\[ V_y = V\sin\theta \]

If the horizontal and vertical components are known, the original vector magnitude can be found using:

\[ V = \sqrt{V_x^2+V_y^2} \]

Its direction can be found using:

\[ \theta = \tan^{-1} \left( \frac{V_y}{V_x} \right) \]
Important: Always check the quadrant when determining the direction of a vector.

15. Projectile Motion

A projectile is an object that moves through the air under the influence of gravity, assuming air resistance is neglected.

Examples include:

  • A basketball after being thrown
  • A soccer ball kicked through the air
  • A stone thrown from a cliff
  • A golf ball
  • A ball rolling off a table

Projectile motion is two-dimensional.

Most Important Idea: Horizontal and vertical motion can be analysed independently.

16. Horizontal Motion of a Projectile

When air resistance is ignored, there is no horizontal acceleration acting on a projectile.

\[ a_x = 0 \]

Therefore, horizontal velocity remains constant:

\[ v_x = \text{constant} \]

Horizontal displacement is calculated using:

\[ \Delta x = v_x t \]

17. Vertical Motion of a Projectile

Vertical motion is affected by Earth's gravitational acceleration. Near Earth's surface:

\[ g = 9.8\,\text{m/s}^2 \]

If upward is selected as positive:

\[ a_y = -9.8\,\text{m/s}^2 \]

If downward is selected as positive:

\[ a_y = +9.8\,\text{m/s}^2 \]

The important thing is to choose one coordinate system and remain consistent throughout the calculation.

Vertical motion can be solved using the standard kinematic equations:

\[ v_{yf}=v_{yi}+a_yt \]
\[ \Delta y = v_{yi}t+\frac{1}{2}a_yt^2 \]

18. Projectile Launched Horizontally

Imagine a ball rolling off the edge of a table.

Initially, the ball has horizontal velocity but no initial vertical velocity:

\[ v_{yi}=0 \]

However, gravity immediately begins accelerating the ball downward.

Horizontal Direction

\[ \Delta x = v_xt \]

Vertical Direction

\[ \Delta y = \frac{1}{2}gt^2 \]

The horizontal and vertical motions occur simultaneously.

The horizontal motion does not determine the time the object takes to fall. The vertical motion determines the fall time, and that same time is then used in the horizontal equation.

19. Projectile Launched at an Angle

Suppose a projectile is launched with an initial speed vi at an angle θ above the horizontal.

The initial velocity must first be separated into horizontal and vertical components.

Horizontal Component

\[ v_{ix}=v_i\cos\theta \]

Vertical Component

\[ v_{iy}=v_i\sin\theta \]

Then solve the two directions independently.

Horizontal Motion

\[ a_x=0 \]
\[ \Delta x=v_{ix}t \]

Vertical Motion

\[ a_y=-9.8\,\text{m/s}^2 \]
\[ v_{yf}=v_{iy}+a_yt \]
\[ \Delta y= v_{iy}t+\frac{1}{2}a_yt^2 \]

20. Maximum Height

At the highest point of a projectile's path, its vertical velocity becomes zero.

\[ v_y=0 \]

However, this does not mean the projectile has completely stopped. Its horizontal velocity can still be present.

\[ v_x\neq0 \]

At maximum height, use the vertical kinematic equations to determine the height or the time taken to reach that point.

For example:

\[ v_f^2=v_i^2+2a\Delta y \]

When solving for maximum height, remember that the relevant initial velocity is the vertical component of the initial velocity.

21. Time of Flight and Range

For an ideal projectile launched and landing at the same vertical height, the total flight time is determined using the vertical motion.

Once the flight time is known, horizontal range can be found using:

\[ R=v_xt \]

For a projectile launched at speed vi at angle θ and landing at the same height, the ideal range is:

\[ R= \frac{v_i^2\sin(2\theta)}{g} \]

Under ideal conditions, the maximum theoretical range occurs at:

\[ \theta=45^\circ \]

Real-world air resistance can change the actual result.

22. Worked Example: Projectile Motion

A ball is launched at 20 m/s at an angle of 30° above the horizontal. Find its initial horizontal and vertical velocity components.

Step 1: Horizontal Component

\[ v_x=v_i\cos\theta \]
\[ v_x=20\cos30^\circ \]
\[ v_x\approx17.3\,\text{m/s} \]

Step 2: Vertical Component

\[ v_y=v_i\sin\theta \]
\[ v_y=20\sin30^\circ \]
\[ v_y=10\,\text{m/s} \]

Therefore, the initial velocity consists of:

\[ \boxed{v_x\approx17.3\,\text{m/s}} \]
\[ \boxed{v_y=10\,\text{m/s}} \]

The horizontal component remains constant while gravity changes the vertical component.

23. Common Kinematics Mistakes

Mistake 1: Confusing Distance and Displacement

Distance is the total path travelled, while displacement is the change from the initial position to the final position.

Mistake 2: Confusing Speed and Velocity

Velocity requires both magnitude and direction.

Mistake 3: Assuming Negative Acceleration Means Slowing Down

Always compare the direction or signs of velocity and acceleration.

Mistake 4: Forgetting Units

Always include appropriate units such as m, s, m/s, and m/s².

Mistake 5: Incorrect Sign for Gravity

If upward is positive, gravitational acceleration is negative:

\[ a_y=-9.8\,\text{m/s}^2 \]

Mistake 6: Treating Projectile Motion as One-Dimensional

Separate projectile motion into horizontal and vertical components.

Mistake 7: Assuming Velocity Is Zero at Maximum Height

Only the vertical component of velocity is zero at maximum height. Horizontal velocity may still exist.

24. A Reliable Kinematics Problem-Solving Method

Use the following method for almost every Grade 11 kinematics problem.

Step 1 — Draw the Situation

Make a simple diagram showing the object, direction of motion, known values, and any relevant angles or distances.

Step 2 — Choose the Positive Direction

For example:

  • Right = positive
  • Up = positive

Step 3 — List the Known Quantities

\[ v_i,\quad v_f,\quad a,\quad t,\quad \Delta x \]

Step 4 — Identify the Unknown

Clearly determine what the question is asking you to calculate.

Step 5 — Choose the Equation

Select an equation that contains the quantities you know and the quantity you need.

Step 6 — Substitute with Correct Signs

For example, if upward is positive:

\[ a=-9.8\,\text{m/s}^2 \]

Step 7 — Calculate

Show your mathematical steps instead of writing only the final answer.

Step 8 — Check the Answer

Ask yourself whether the magnitude, direction, and units make physical sense.

25. Kinematics Formula Sheet

Average Velocity

\[ v_{\text{avg}}= \frac{\Delta x}{\Delta t} \]

Average Acceleration

\[ a_{\text{avg}}= \frac{\Delta v}{\Delta t} \]

First Kinematic Equation

\[ v_f=v_i+at \]

Second Kinematic Equation

\[ \Delta x= v_it+\frac{1}{2}at^2 \]

Third Kinematic Equation

\[ v_f^2= v_i^2+2a\Delta x \]

Fourth Kinematic Equation

\[ \Delta x= \frac{v_i+v_f}{2}t \]

Projectile Components

\[ v_x=v_i\cos\theta \]
\[ v_y=v_i\sin\theta \]

Horizontal Projectile Motion

\[ \Delta x=v_xt \]

Vertical Projectile Motion

\[ \Delta y= v_{yi}t+\frac{1}{2}a_yt^2 \]

Acceleration Due to Gravity

\[ g=9.8\,\text{m/s}^2 \]

26. Quick Revision Checklist

Before your Grade 11 Physics test, make sure you can:

  • Define kinematics.
  • Distinguish distance from displacement.
  • Distinguish scalar from vector quantities.
  • Distinguish speed from velocity.
  • Calculate average velocity.
  • Calculate acceleration.
  • Interpret positive and negative signs.
  • Identify uniform and non-uniform motion.
  • Use the four kinematic equations.
  • Interpret position-time graphs.
  • Interpret velocity-time graphs.
  • Interpret acceleration-time graphs.
  • Calculate vector components.
  • Resolve vectors into horizontal and vertical components.
  • Explain two-dimensional motion.
  • Separate projectile motion into horizontal and vertical components.
  • Use \(g=9.8\,\text{m/s}^2\).
  • Find projectile time, height, and range.
  • Recognize that \(v_y=0\) at maximum height.
  • Solve problems systematically using diagrams and signs.

27. Rahul Sir's Final Revision Tip

Kinematics becomes much easier when you stop memorizing isolated formulas and start understanding the relationships between physical quantities.

Ask yourself:

  • Where is the object? → Position and displacement
  • How far did it travel? → Distance
  • How quickly is it moving? → Speed
  • In which direction is it moving? → Velocity
  • How is its velocity changing? → Acceleration

For two-dimensional motion, remember one powerful strategy:

Break the motion into independent horizontal and vertical components, solve each component, and combine the results when necessary.

If you understand vectors, graphs, acceleration, the four kinematic equations, and projectile components, you have mastered the essential foundation of Grade 11 Kinematics.

Keep your diagrams clear, choose your coordinate system carefully, maintain consistent signs, show your calculations, and always include units in your final answer.

Keep practising — Physics becomes easier when you understand the motion, not just the formula.

— Rahul Sir | ODTutor Canada