- Position & Displacement
- Average Velocity
- Acceleration
- Equations of Motion
- Graphs
- Relative Velocity
- Summary
Welcome, dear student! In Unit 1, we studied vectors. Now we enter Unit 2: Uniformly Accelerated Motion. This is one of the most important units in Grade 10 Physics. In Grade 9, you learned about uniform motion, which means motion at constant speed in a straight line. But in real life, things rarely move at a steady speed. A bus speeds up from a stop, a car slows down at a red light, a ball falls faster and faster. All of these are examples of accelerated motion. In this unit, we will learn how to describe and calculate this type of motion step by step.
This unit has six sections. We will cover every section deeply with examples from your textbook and plenty of exam-style questions. Let us start!
2.1 Position and Displacement
Before we describe how fast something moves, we first need to know where it is. That brings us to the idea of position. Let me explain this carefully because many students confuse position with distance, and this confusion costs marks in exams.
What is Position?
Imagine you are standing on a road. If someone calls you and asks “Where are you?”, you might say “I am 500 metres from the school gate.” You just described your position. Position is the location of an object compared to a known reference point.
But saying “I am here” is useless. You must always describe position relative to a reference point. In physics, we call this reference point the origin. The set of axes you use to measure position is called a frame of reference.
Position is the location of an object with reference to an origin. It can be positive or negative. Its SI unit is metre (m).
Think about it practically. If the new school is 300 m from Kemal’s house, then Kemal’s house is the origin. The school’s position is +300 m (if we choose the direction from Kemal’s house to the school as positive). Position always needs a number, a unit, and a reference point.
What is Displacement?
When an object moves, its position changes. The measure of this change is called displacement. Displacement is a vector quantity from Unit 1 because it has both magnitude and direction.
Displacement is the difference between the final position and the initial position of an object. It is the shortest straight-line distance from start to finish, with direction. It is a vector quantity.
$$ \Delta s = s – s_0 $$
Δs = displacement | s = final position | s0 = initial position
The symbol Δ (delta) means “change in.” So Δs means “change in position.” Displacement can be positive, negative, or zero. It depends on direction relative to your chosen positive direction.
Distance vs. Displacement
This is one of the most tested differences in Ethiopian exams. You must know it perfectly.
| Feature | Distance | Displacement |
|---|---|---|
| Type | Scalar (magnitude only) | Vector (magnitude + direction) |
| What it measures | Total path length travelled | Straight line from start to end |
| Can it be zero? | Never (if object moved) | Yes (if start = end position) |
| Which is larger? | Distance is always greater than or equal to |Displacement| | |Displacement| is always less than or equal to Distance |
Now consider the object goes from O to A (60 km) and then all the way back to O. The distance is 60 + 60 = 120 km. But the displacement is 60 – 60 = 0 km. The object is back where it started! This is a very important idea.
Problem: A person walks 70 m East, and then 30 m West. Find the displacement.
Let East be positive.
Final position: s = +70 + (-30) = +40 m
Initial position: s0 = 0 m
$$ \Delta s = s – s_0 = 40 – 0 = \textbf{40 m East} $$
Displacement = 40 m East. But total distance = 70 + 30 = 100 m. Distance is greater than displacement!
The magnitude of displacement can NEVER exceed the total distance. This is a fundamental rule. If a question gives you a round trip, displacement = 0 but distance is not zero. Many MCQs test this exact point.
Practice Questions — Position and Displacement
Explanation: Distance = 15 + 20 = 35 m (total path). For displacement, the two parts are at 90 degrees, so by Pythagoras: |Δs| = √(15² + 20²) = √(225 + 400) = √625 = 25 m. Textbook review question.
Explanation: The 8 m North and 8 m South cancel each other (net N-S = 0). Only 5 m East remains. Displacement = 5 m East. Distance would be 8+8+5 = 21 m, but displacement considers direction and cancellation.
Explanation: From your textbook Exercise 2.3: the shortest path between two points is a straight line (displacement). The actual path (distance) is always equal to or longer. This rule can never be broken.
2.2 Average Velocity and Instantaneous Velocity
Now we know how to describe position and displacement. The next question is: how fast does an object move? That is what velocity tells us. But there are two types you must understand clearly.
What is Average Velocity?
When a car travels on a road, it does not maintain the same speed throughout. Sometimes it goes faster, sometimes slower, sometimes it stops. To describe the overall motion, we use average velocity.
Average velocity is the total displacement divided by the total time taken. It is a vector quantity.
$$ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{s – s_0}{t – t_0} $$
v̄ = average velocity (m/s) | Δs = displacement (m) | Δt = time interval (s)
Notice: the numerator is displacement, NOT distance. This is very important for exams. The SI unit is metres per second (m/s). Other units are km/h, cm/s.
What is Average Speed?
Do not confuse average velocity with average speed!
$$ \text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} $$
Average velocity is not equal to Average speed (unless moving in one direction in a straight line).
Velocity uses displacement. Speed uses distance. Read the question carefully!
Problem: It takes you 10 minutes to walk with an average velocity of 1.2 m/s North from the bus stop to the museum. What is your displacement?
Given: Δt = 10 min = 600 s, v̄ = 1.2 m/s North
Rearrange: Δs = v̄ × Δt
$$ \Delta s = 1.2 \times 600 = \textbf{720 m, North} $$
Problem: A passenger in a bus took 8 s to move 4 m forward. What is his average velocity?
$$ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{4}{8} = \textbf{0.5 m/s} $$
Problem: A car travels at 50 km/h for 100 km, then at 100 km/h for another 100 km. What is the average speed for the 200 km trip?
Step 1: Find time for each part:
$$ t_1 = \frac{100}{50} = 2 \text{ h}, \quad t_2 = \frac{100}{100} = 1 \text{ h} $$
Step 2: Total distance = 200 km, Total time = 3 h
$$ \text{Average speed} = \frac{200}{3} = \textbf{66.7 km/h} $$
In Example 2.4, many students wrongly calculate (50 + 100) / 2 = 75 km/h. This is WRONG! You cannot simply average two speeds. The car spent more time at 50 km/h, so the true average must be closer to 50 than to 100. Always use: Average speed = Total distance ÷ Total time. Find each time separately first.
Instantaneous Velocity
If your average velocity for a trip is 20 m/s, were you going exactly 20 m/s at every moment? No. Sometimes faster, sometimes slower. Instantaneous velocity is the velocity at one specific instant.
Instantaneous velocity is the velocity of an object at a specific instant in time. It is the rate of change of displacement as the time interval approaches zero. The magnitude of instantaneous velocity is what a speedometer shows.
$$ v = \frac{s – s_0}{t – t_0} \quad \text{when} \quad (t – t_0) \rightarrow 0 $$
Average velocity = Instantaneous velocity only when the object moves with constant velocity (no speed changes, no direction changes). In all other cases, they are different.
Practice Questions — Average and Instantaneous Velocity
Explanation: t1 = 120/60 = 2 h, t2 = 120/40 = 3 h. Total = 240 km in 5 h. Average speed = 240/5 = 48 km/h. NOT 50 km/h (the simple average). Same method as Example 2.4.
Explanation: Displacement = 3 – 3 = 0 km. Average velocity = 0 / 45 min = 0. Note: average SPEED = 6/45 = 0.133 km/min. But the question asks for velocity. Classic exam trick!
Explanation: The speedometer shows a number at one instant with no direction. That is instantaneous speed (magnitude of instantaneous velocity). Not velocity because no direction is shown. From textbook Exercise 2.5.
2.3 Acceleration
Now we come to the heart of this unit. When an object’s velocity changes, we say it is accelerating. But “acceleration” does not always mean “speeding up.” Let me explain this very carefully because this is where many students get confused.
What is Acceleration?
Think about a bus at a stop. When the driver starts the bus, the velocity increases from 0 to some value. The bus is accelerating. Now think about a bus that is moving and the driver applies the brakes. The velocity decreases. The bus is also accelerating! In physics, any change in velocity — whether speeding up, slowing down, or changing direction — is called acceleration.
Acceleration is the rate of change of velocity with respect to time. It tells us how quickly the velocity is changing. It is a vector quantity (it has direction).
$$ a = \frac{\Delta v}{\Delta t} = \frac{v – v_0}{t – t_0} $$
a = acceleration (m/s²) | v = final velocity | v0 = initial velocity
Δv = change in velocity | Δt = time interval
The SI unit of acceleration is metres per second squared (m/s²). This unit makes sense if you think about it: velocity is m/s, time is s, so (m/s) / s = m/s². It tells you how many m/s the velocity changes by each second.
Understanding Positive and Negative Acceleration
This is very important for exams. Let me explain with clear examples.
Positive acceleration: Velocity is increasing in the positive direction. The object speeds up. Example: a car going East goes from 10 m/s to 20 m/s.
Negative acceleration (deceleration): Velocity is decreasing. The object slows down. Example: a car going East goes from 20 m/s to 10 m/s.
Important: Whether acceleration is positive or negative depends on your chosen positive direction AND whether the object is speeding up or slowing down.
Problem: A car moving East increases its velocity from 10 m/s to 30 m/s in 5 seconds. Find the acceleration.
v0 = 10 m/s, v = 30 m/s, t = 5 s
$$ a = \frac{v – v_0}{t} = \frac{30 – 10}{5} = \frac{20}{5} = \textbf{+4 m/s}^2 $$
Positive acceleration means the car is speeding up in the East direction. The velocity increases by 4 m/s every second.
Problem: A car moving East slows down from 25 m/s to 5 m/s in 4 seconds. Find the acceleration.
v0 = 25 m/s, v = 5 m/s, t = 4 s (taking East as positive)
$$ a = \frac{5 – 25}{4} = \frac{-20}{4} = \textbf{-5 m/s}^2 $$
The negative sign means the car is decelerating (slowing down). The velocity decreases by 5 m/s every second.
What is Uniformly Accelerated Motion?
Uniformly accelerated motion is motion in which the acceleration is constant — it does not change with time. The velocity changes by equal amounts in equal time intervals. For example, if a car has an acceleration of 2 m/s², its velocity increases by 2 m/s every second: 0, 2, 4, 6, 8, 10… This is a steady, predictable change.
- Acceleration = rate of change of velocity
- Formula: a = (v – v0) / t
- SI unit: m/s²
- Acceleration is a vector — direction matters
- Positive acceleration = speeding up (in chosen positive direction)
- Negative acceleration = slowing down (deceleration)
- Uniform acceleration = acceleration is constant (does not change)
- A turning object also accelerates (direction of velocity changes)
Practice Questions — Acceleration
Explanation: a = (v – v0) / t = (15 – 5) / 5 = 10 / 5 = 2 m/s². The velocity changes by 2 m/s each second. Simple and direct application of the formula.
Explanation: Final velocity v = 0 m/s (car stops), v0 = 20 m/s. a = (0 – 20) / 4 = -20/4 = -5 m/s². The negative sign shows deceleration (slowing down). Option A is wrong because it ignores the sign. Option D is wrong because it multiplies instead of dividing.
Explanation: The negative sign in acceleration tells us the velocity is decreasing (when moving in the positive direction). The magnitude 3 m/s² tells us the rate: the velocity decreases by 3 m/s each second. Option B is partially true but less precise than C. Option A is wrong because a negative acceleration does not necessarily mean the object moves in the negative direction.
2.4 Equations of Motion with Constant Acceleration
This is the most important section of the entire unit. When an object moves with uniform (constant) acceleration, we can use four special equations to solve problems. These equations connect five variables: displacement, initial velocity, final velocity, acceleration, and time. If you know any three of these, you can find the other two. Memorise these four equations — they will appear in every physics exam you take from Grade 10 to university!
The Five Variables
Before writing the equations, let me list the five variables clearly. Every problem in this section uses these symbols:
| Symbol | Meaning | SI Unit |
|---|---|---|
| v0 | Initial velocity (velocity at t = 0) | m/s |
| v | Final velocity (velocity at time t) | m/s |
| a | Acceleration (constant) | m/s² |
| t | Time interval | s |
| Δs | Displacement | m |
The Four Equations of Motion
| Equation | What It Finds | Variables Used | No Time? |
|---|---|---|---|
| $$ v = v_0 + at $$ | Final velocity | v, v0, a, t | NO |
| $$ \Delta s = v_0 t + \frac{1}{2}at^2 $$ | Displacement | Δs, v0, a, t | NO |
| $$ v^2 = v_0^2 + 2a\Delta s $$ | Final velocity (no t) | v, v0, a, Δs | YES |
| $$ \Delta s = \frac{1}{2}(v_0 + v)t $$ | Displacement (no a) | Δs, v0, v, t | NO |
Let me now explain each equation deeply with examples.
Equation 1: v = v0 + at
This is the simplest equation. It tells you the final velocity when you know the initial velocity, acceleration, and time. You can also rearrange it to find acceleration (a = (v – v0) / t) or time (t = (v – v0) / a).
Problem: A car starts from rest and accelerates uniformly at 3 m/s². What is its velocity after 6 seconds?
Given: v0 = 0 (starts from rest), a = 3 m/s², t = 6 s
Find: v
$$ v = v_0 + at = 0 + 3 \times 6 = \textbf{18 m/s} $$
“Starts from rest” always means v0 = 0. Remember this — it appears in many exam questions!
Equation 2: Δs = v0t + ½at²
This equation finds the displacement. It accounts for both the distance covered due to the initial velocity (v0t) and the extra distance due to acceleration (½at²). If the object starts from rest (v0 = 0), the equation simplifies to Δs = ½at².
Problem: A car starts from rest with acceleration of 2 m/s². How far does it travel in 10 seconds?
Given: v0 = 0, a = 2 m/s², t = 10 s
$$ \Delta s = 0 \times 10 + \frac{1}{2} \times 2 \times 10^2 = 0 + 100 = \textbf{100 m} $$
Equation 3: v² = v0² + 2aΔs
This equation is special because it does not contain time. Use it when the question does not give you time and does not ask for time. It connects final velocity, initial velocity, acceleration, and displacement directly.
Problem: A car moving at 15 m/s accelerates at 2 m/s² over a distance of 100 m. What is its final velocity?
Given: v0 = 15 m/s, a = 2 m/s², Δs = 100 m. No time given!
$$ v^2 = 15^2 + 2 \times 2 \times 100 = 225 + 400 = 625 $$
$$ v = \sqrt{625} = \textbf{25 m/s} $$
Since velocity is a vector, v = +25 m/s (same direction as initial velocity). Whenever you take a square root in physics, consider whether the answer should be positive or negative.
Equation 4: Δs = ½(v0 + v)t
This equation does not contain acceleration. It says that displacement equals the average of initial and final velocities multiplied by time. It is very useful when you know both velocities and time but not acceleration.
Problem: A train accelerates uniformly from 10 m/s to 30 m/s in 8 seconds. Find the displacement.
Given: v0 = 10 m/s, v = 30 m/s, t = 8 s
$$ \Delta s = \frac{1}{2}(10 + 30) \times 8 = \frac{1}{2} \times 40 \times 8 = \textbf{160 m} $$
Follow this simple method for every problem:
Step 1: Write down what you are given (three values)
Step 2: Write down what you need to find
Step 3: Pick the equation that contains exactly those four variables
Step 4: If time is NOT given and NOT asked for, use Equation 3 (v² = v0² + 2aΔs)
Step 5: If acceleration is NOT given and NOT asked for, use Equation 4
Step 6: If object “starts from rest,” v0 = 0 automatically
These four equations ONLY work for uniform (constant) acceleration. If acceleration is not constant, you cannot use these equations. In Grade 10, all problems will have constant acceleration unless stated otherwise.
Practice Questions — Equations of Motion
Explanation: v0 = 0 (rest), a = 5 m/s², t = 4 s. Use Δs = ½at² = ½ × 5 × 16 = 40 m. You could also find v = 0 + 5×4 = 20 m/s, then use Δs = ½(0+20)×4 = 40 m. Both methods give the same answer!
Explanation: At the highest point, v = 0. Take upward as positive, so a = -10 m/s². v0 = 20 m/s. No time given, so use v² = v0² + 2aΔs: 0 = 400 + 2(-10)Δs → 0 = 400 – 20Δs → Δs = 400/20 = 20 m. This is a very common type of exam question!
Explanation: “Stops” means v = 0. v0 = 20 m/s, a = -4 m/s². Use v = v0 + at: 0 = 20 + (-4)t → 4t = 20 → t = 5 s. “Decelerates at 4 m/s²” means acceleration is -4 m/s² if the car moves in the positive direction.
2.5 Graphical Representation of Uniformly Accelerated Motion
In physics, we often represent motion using graphs. Graphs give us a visual picture of how position, velocity, and acceleration change with time. For your exam, you need to understand three types of graphs: position-time graphs, velocity-time graphs, and acceleration-time graphs. The velocity-time graph is the most important one for uniformly accelerated motion.
Position-Time Graph for Uniformly Accelerated Motion
When an object has uniform acceleration, its position changes faster and faster as time passes. On a position-time graph, this produces a curved line (a parabola).
The slope of a position-time graph at any point equals the instantaneous velocity at that time. Since the curve gets steeper, the slope increases, which means the velocity is increasing — that is acceleration!
Straight line on s-t graph → constant velocity (uniform motion, no acceleration)
Curved line on s-t graph → changing velocity (accelerated motion)
Horizontal line on s-t graph → object at rest (zero velocity)
Velocity-Time Graph for Uniformly Accelerated Motion
This is the most important graph for this unit. When acceleration is constant, the velocity changes at a steady rate, so the velocity-time graph is a straight line with a constant slope.
Rule 1: The slope of a v-t graph = acceleration
$$ \text{Slope} = \frac{\Delta v}{\Delta t} = a $$
Rule 2: The area under a v-t graph = displacement
$$ \text{Area} = \Delta s $$
Let me explain both rules with examples.
Finding Acceleration from the Slope
If the v-t graph is a straight line going upward, the slope is positive — positive acceleration (speeding up). If it goes downward, the slope is negative — negative acceleration (slowing down). If it is horizontal (flat), the slope is zero — no acceleration (constant velocity).
Problem: A v-t graph shows a straight line from (0, 5) to (4, 25) on the v-t axes (v in m/s, t in s). Find the acceleration.
$$ a = \text{Slope} = \frac{v_2 – v_1}{t_2 – t_1} = \frac{25 – 5}{4 – 0} = \frac{20}{4} = \textbf{5 m/s}^2 $$
The positive slope confirms the object is speeding up.
Finding Displacement from the Area
The area under the v-t graph between two time values gives the displacement during that time. For a straight-line v-t graph, the area is a trapezoid (or triangle if starting from rest).
If the line does not start from zero, the area is a trapezoid:
If starting from rest: Δs = ½ × base × height = ½ × t × v
If starting from v0: Δs = ½ × (v0 + v) × t
(This is the same as Equation 4! The graph gives you the equation.)
Acceleration-Time Graph
For uniformly accelerated motion, acceleration is constant. So the acceleration-time graph is a horizontal line at the value of a.
The area under an a-t graph gives the change in velocity (Δv). Since the line is horizontal, the area is just a rectangle: Δv = a × t, which matches v = v0 + at when v0 = 0.
- s-t graph: Slope = velocity. Curved line = acceleration. Straight line = constant velocity.
- v-t graph: Slope = acceleration. Area under curve = displacement. Straight line = constant acceleration.
- a-t graph: Area under curve = change in velocity. Horizontal line = constant acceleration.
Practice Questions — Graphs of Motion
Explanation: A horizontal line on a v-t graph means velocity is not changing (constant). Slope = Δv/Δt = 0, so acceleration = 0. The object moves with uniform motion (constant velocity). It is NOT at rest unless the horizontal line is at v = 0.
Explanation: The area under a v-t graph represents displacement, not distance. If the graph goes below the time axis (negative velocity), the area below counts as negative, and the total area gives displacement (not total distance). Only when the graph stays entirely above the axis does displacement equal distance.
Explanation: On an s-t graph, the slope = velocity. If the curve gets steeper, the slope is increasing, meaning velocity is increasing. Increasing velocity means positive acceleration. A straight line would mean constant velocity (no acceleration), and a flattening curve would mean deceleration.
Explanation: Slope of v-t graph = acceleration. Negative slope = negative acceleration = deceleration (velocity is decreasing). The object could still be moving forward (positive velocity) but slowing down. The sign of the slope tells us about acceleration, not about the direction of motion itself.
2.6 Relative Velocity in One Dimension
Here is an interesting question: if you are sitting in a bus that is moving at 60 km/h, and you throw a ball forward at 10 km/h (relative to the bus), how fast is the ball moving relative to the ground? The answer is 70 km/h. This is the idea of relative velocity — velocity depends on who is observing it.
What is Relative Velocity?
The velocity of an object is not absolute. It depends on the frame of reference of the observer. A passenger walking forward in a moving train has one velocity relative to the train, but a different velocity relative to the ground.
Relative velocity is the velocity of one object as observed from another moving object. It tells you how fast one object appears to be moving when you are watching from a different moving object.
$$ v_{AB} = v_A – v_B $$
vAB = velocity of A relative to B
vA = velocity of A relative to the ground
vB = velocity of B relative to the ground
This formula says: to find how fast A appears to move when observed from B, subtract B’s velocity from A’s velocity. Let me explain with clear examples.
Objects Moving in the Same Direction
Problem: Car A moves at 80 km/h East. Car B moves at 60 km/h East. What is the velocity of Car A relative to Car B?
$$ v_{AB} = v_A – v_B = 80 – 60 = \textbf{20 km/h East} $$
To a person in Car B, Car A appears to be moving away at only 20 km/h, even though Car A is actually going 80 km/h relative to the ground. This makes sense — if you are driving at 60 km/h and another car passes you at 80 km/h, it seems like it is only going 20 km/h faster than you.
Objects Moving in Opposite Directions
Problem: Car A moves at 50 km/h East. Car B moves at 40 km/h West. What is the velocity of Car A relative to Car B?
Let East be positive. So vA = +50 km/h, vB = -40 km/h.
$$ v_{AB} = 50 – (-40) = 50 + 40 = \textbf{90 km/h East} $$
When objects move in opposite directions, their relative velocity is the SUM of their speeds. This is why head-on collisions are so dangerous — the relative speed is very high!
Practical Applications
Relative velocity is very important in real life. Here are some examples:
- Overtaking: When you overtake a slower vehicle, your relative velocity tells you how quickly you will pass it. If you are going 70 km/h and the truck ahead is going 50 km/h, you approach it at only 20 km/h relative speed.
- Head-on danger: Two cars moving towards each other at 60 km/h each have a relative velocity of 120 km/h. This is why opposite-direction accidents are far more deadly.
- River crossing: A boat crossing a river must account for the river current’s velocity relative to the ground.
- Relative velocity = vA – vB (always subtract the observer’s velocity)
- Same direction: relative velocity = difference of speeds
- Opposite directions: relative velocity = sum of speeds
- If vAB = 0, both objects move at the same velocity (they stay at the same distance from each other)
Practice Questions — Relative Velocity
Explanation: vAB = vA – vB = 25 – 15 = 10 m/s. Positive means same direction as the original motion. To a passenger on Train B, Train A appears to move forward at 10 m/s. Same direction means you subtract the speeds.
Explanation: Let North be positive. vcar = +30 m/s, vtruck = -20 m/s (South is negative). vcar, truck = 30 – (-20) = 30 + 20 = 50 m/s North. Opposite directions means you ADD the speeds. To the truck driver, the car appears to approach at 50 m/s!
Explanation: vAB = 20 – 20 = 0 m/s. If both cars move at the same speed in the same direction, each appears stationary relative to the other. They maintain the same distance. This is why on a highway, cars moving at the same speed appear to not move relative to each other.
Complete Unit Summary — Exam Preparation
Excellent work, dear student! You have now completed the entire Unit 2 on Uniformly Accelerated Motion. Here is everything you need to remember before your exam.
Displacement: $$ \Delta s = s – s_0 $$
Average Velocity: $$ \bar{v} = \frac{\Delta s}{\Delta t} $$
Average Speed: $$ \text{Avg speed} = \frac{\text{Total distance}}{\text{Total time}} $$
Acceleration: $$ a = \frac{v – v_0}{t} $$
Equation 1: $$ v = v_0 + at $$
Equation 2: $$ \Delta s = v_0 t + \frac{1}{2}at^2 $$
Equation 3: $$ v^2 = v_0^2 + 2a\Delta s $$
Equation 4: $$ \Delta s = \frac{1}{2}(v_0 + v)t $$
Relative Velocity: $$ v_{AB} = v_A – v_B $$
- Position = location relative to origin (can be + or -)
- Distance = total path length (scalar, always ≥ 0)
- Displacement = final minus initial position (vector, can be 0)
- Distance is always ≥ |Displacement| — always true
- Average velocity uses displacement; average speed uses distance
- NEVER simply average two speeds — use total distance / total time
- Instantaneous velocity = velocity at one instant (speedometer shows instantaneous speed)
- Acceleration = rate of change of velocity; unit = m/s²
- Positive acceleration = speeding up; Negative = slowing down (deceleration)
- “Starts from rest” always means v0 = 0
- “Comes to rest” or “stops” means v = 0
- Four equations of motion work ONLY for constant acceleration
- Use Equation 3 (v² = v0² + 2aΔs) when time is not involved
- v-t graph: slope = acceleration, area = displacement
- s-t graph: slope = velocity, curved = acceleration
- Relative velocity: same direction = subtract, opposite direction = add
✓ Always check: does the question ask for velocity or speed, distance or displacement?
✓ Write down “Given, Find, Formula” for every numerical problem
✓ Convert units before calculating: minutes to seconds, km/h to m/s
✓ 1 km/h = 5/18 m/s — memorise this conversion
✓ For free-fall problems: take upward as positive, so g = -10 m/s²
✓ “Starts from rest” means v0 = 0; “Stops” means v = 0
✓ When time is not given and not asked, use v² = v0² + 2aΔs
✓ For average speed with two different speeds, find each time first
✓ Negative acceleration means slowing down only if velocity is positive
✓ On v-t graphs: area ABOVE the axis = positive displacement, BELOW = negative
This unit is the foundation for all of mechanics in Grade 11 and Grade 12. If you understand these concepts well now, you will have a much easier time later. Practise as many problems as you can, especially using the four equations of motion. The more you practise, the faster you will recognise which equation to use. Good luck with your exam!