Rectilinear Acceleration and Equations of Motion for SS 1

Rectilinear Acceleration

Performance Objectives

By the end of the lesson, students should be able to:

1. define acceleration and state the unit;
2. explain the concept of uniform motion;
3. derive the formula of acceleration;
4. plot velocity-time graph;
5. deduce acceleration from velocity-time graph.

Understanding Motion: A Quick Review

Before discussing acceleration, it is important to recall some basic concepts of motion. Motion describes how an object changes its position over time. When an object moves, its position changes relative to a fixed point or reference.

Speed is the rate at which an object covers distance. It is a scalar quantity, meaning it only has magnitude. For example, a car moving at 60 kilometres per hour (km/h) has a speed of 60 km/h.

Velocity is the rate at which an object changes its displacement. It is a vector quantity, meaning it has both magnitude and direction. For example, a car moving at 60 km/h towards the East has a velocity of 60 km/h East.

In rectilinear motion, objects move along a straight line. This unit focuses on motion in a straight line, where the direction is usually constant or simply reverses along the line.

Meaning of Acceleration

Acceleration is the rate of change of velocity of an object.

When an object’s velocity changes, it is said to be accelerating. This change can be an increase in speed, a decrease in speed, or a change in the direction of motion. Since this unit focuses on rectilinear motion, the primary changes in velocity will be changes in speed along a straight line.

Acceleration is a vector quantity, meaning it has both magnitude and direction. Its direction is the same as the direction of the change in velocity.

The unit of acceleration is metres per second squared (m/s²).

This unit comes from dividing a change in velocity (measured in m/s) by the time taken for that change (measured in s). So, (m/s) / s = m/s².

Examples of acceleration in everyday life include:

1. A car speeding up from a traffic light.
2. A bicycle slowing down as it approaches a stop sign.
3. A ball thrown upwards, slowing down as it rises and speeding up as it falls.
4. A sprinter increasing their speed at the start of a race.

Positive and Negative Acceleration

When an object’s velocity increases in the direction of motion, it experiences positive acceleration. For example, a car moving forward and increasing its speed has positive acceleration.

When an object’s velocity decreases in the direction of motion, it experiences negative acceleration. This is also known as deceleration or retardation. For example, a car moving forward and applying brakes to slow down has negative acceleration (or deceleration).

Deriving the Formula for Acceleration

To calculate acceleration, the change in velocity is divided by the time taken for that change.

Consider an object moving in a straight line:

1. Let its initial velocity be ‘u’ (velocity at the start of observation).
2. Let its final velocity be ‘v’ (velocity at the end of observation).
3. Let the time taken for this change in velocity be ‘t’.

The change in velocity is the final velocity minus the initial velocity, which is (v – u).

Therefore, the formula for acceleration (a) is:

a = (v – u) / t

Where:

1. a = acceleration (m/s²)
2. v = final velocity (m/s)
3. u = initial velocity (m/s)
4. t = time taken (s)

Worked Example 1

A car starts from rest and reaches a velocity of 20 m/s in 5 seconds. Calculate its acceleration.

Solution:

Initial velocity (u) = 0 m/s (since it starts from rest)

Final velocity (v) = 20 m/s

Time (t) = 5 s

Using the formula: a = (v – u) / t

a = (20 m/s – 0 m/s) / 5 s

a = 20 m/s / 5 s

a = 4 m/s²

The acceleration of the car is 4 m/s².

Worked Example 2

A motorcycle travelling at 30 m/s applies its brakes and slows down to 10 m/s in 4 seconds. Calculate its acceleration.

Solution:

Initial velocity (u) = 30 m/s

Final velocity (v) = 10 m/s

Time (t) = 4 s

Using the formula: a = (v – u) / t

a = (10 m/s – 30 m/s) / 4 s

a = -20 m/s / 4 s

a = -5 m/s²

The acceleration of the motorcycle is -5 m/s². The negative sign indicates that the motorcycle is decelerating or retarding.

Uniform Motion

The term “uniform motion” can refer to two related concepts: uniform velocity and uniform acceleration.

Uniform Velocity

An object is said to be in uniform velocity when it moves with a constant speed in a constant direction.

In rectilinear motion, uniform velocity means the object is moving at a steady speed in a straight line without changing direction. When an object has uniform velocity, its acceleration is zero because there is no change in velocity.

Examples of uniform velocity:

1. A car moving at a steady 50 km/h on a straight, empty road.
2. A train moving at a constant speed along a straight track.

Uniform Acceleration

An object is said to be in uniform acceleration when its velocity changes by equal amounts in equal intervals of time.

This means the rate of change of velocity is constant. For example, if an object’s velocity increases by 2 m/s every second, it has a uniform acceleration of 2 m/s².

Examples of uniform acceleration:

1. An object falling freely under gravity (ignoring air resistance) experiences uniform acceleration due to gravity.
2. A car accelerating steadily from rest on a straight road.

Velocity-Time Graphs

A velocity-time graph is a graph that shows how the velocity of an object changes over a period of time.

These graphs are powerful tools for analysing motion because they can be used to determine both the acceleration and the displacement of an object.

Plotting Velocity-Time Graphs

To plot a velocity-time graph, follow these steps:

1. Draw the Axes: Draw two perpendicular axes. The horizontal axis (x-axis) represents time (t), usually in seconds (s). The vertical axis (y-axis) represents velocity (v), usually in metres per second (m/s).
2. Label the Axes: Clearly label each axis with the quantity it represents and its unit (e.g., “Time (s)” and “Velocity (m/s)”).
3. Choose a Scale: Select appropriate scales for both axes so that all data points can fit on the graph sheet and the graph is large enough to be easily read.
4. Plot the Data Points: For each pair of time and velocity values, mark a point on the graph.
5. Draw the Graph Line: Connect the plotted points with a smooth line or curve. For uniform acceleration, this will be a straight line.

Worked Example 3: Plotting a Velocity-Time Graph

A cyclist starts from rest and accelerates uniformly. The table below shows the cyclist’s velocity at different times:

Plot a velocity-time graph for the cyclist’s motion.

Solution:

Using graph paper, draw the time axis horizontally and the velocity axis vertically. Plot the points (0,0), (2,4), (4,8), (6,12), and (8,16). Connect these points with a straight line, as shown in the diagram above.

Interpreting Velocity-Time Graphs

The shape of a velocity-time graph tells a lot about the motion of an object:

1. Horizontal Line: A horizontal line on a velocity-time graph indicates that the velocity is constant. This means the object is moving with uniform velocity, and its acceleration is zero.
2. Straight Line Sloping Upwards: A straight line sloping upwards indicates that the velocity is increasing uniformly. This means the object is undergoing uniform positive acceleration.
3. Straight Line Sloping Downwards: A straight line sloping downwards indicates that the velocity is decreasing uniformly. This means the object is undergoing uniform negative acceleration (deceleration or retardation).
4. Curved Line: A curved line indicates that the velocity is changing at a non-uniform rate. This means the object is undergoing non-uniform acceleration.

Deducing Acceleration from Velocity-Time Graphs

The acceleration of an object can be deduced from the gradient (slope) of its velocity-time graph.

The gradient of a line on a graph is calculated as the change in the vertical axis value divided by the change in the horizontal axis value.

For a velocity-time graph:

Gradient = (Change in Velocity) / (Change in Time)

Since acceleration is defined as the rate of change of velocity, the gradient of the velocity-time graph directly gives the acceleration.

a = (v₂ – v₁) / (t₂ – t₁)

Where (t₁, v₁) and (t₂, v₂) are any two points on the straight line segment of the graph.

Worked Example 4: Deducing Acceleration from a Graph

Using the velocity-time graph from Worked Example 3 (for the cyclist), deduce the acceleration of the cyclist.

Solution:

To find the acceleration, calculate the gradient of the straight line. Choose two points on the line, for example:

Point 1: (t₁ = 2 s, v₁ = 4 m/s)

Point 2: (t₂ = 8 s, v₂ = 16 m/s)

Change in Velocity (Δv) = v₂ – v₁ = 16 m/s – 4 m/s = 12 m/s

Change in Time (Δt) = t₂ – t₁ = 8 s – 2 s = 6 s

Acceleration (a) = Δv / Δt = 12 m/s / 6 s = 2 m/s²

The acceleration of the cyclist is 2 m/s².

Calculating Displacement from Velocity-Time Graphs

The area under a velocity-time graph represents the displacement of the object.

For rectilinear motion, if the velocity is always positive, the area represents the total distance covered. If the velocity becomes negative, the area below the time axis represents displacement in the opposite direction.

To calculate the area, the shape under the graph is often divided into simpler geometric shapes like rectangles, triangles, or trapeziums.

1. For a rectangle (uniform velocity): Area = length × width = velocity × time.
2. For a triangle (uniform acceleration from rest): Area = ½ × base × height = ½ × time × final velocity.
3. For a trapezium (uniform acceleration with initial velocity): Area = ½ × (sum of parallel sides) × height = ½ × (initial velocity + final velocity) × time.

Worked Example 5: Calculating Displacement from a Graph

A car moves with a constant velocity of 15 m/s for 10 seconds. Calculate the displacement using a velocity-time graph.

Solution:

The graph for this motion would be a horizontal line at v = 15 m/s from t = 0 s to t = 10 s. This forms a rectangle.

Area = velocity × time

Area = 15 m/s × 10 s

Area = 150 m

The displacement of the car is 150 m.

Worked Example 6: Calculating Displacement for Uniform Acceleration

A bus starts from rest and accelerates uniformly to a velocity of 20 m/s in 8 seconds. Calculate the displacement of the bus.

Solution:

The graph for this motion would be a straight line from (0,0) to (8 s, 20 m/s). This forms a triangle.

Area = ½ × base × height

Area = ½ × time × final velocity

Area = ½ × 8 s × 20 m/s

Area = 80 m

The displacement of the bus is 80 m.

Worked Example 7: Calculating Displacement for a More Complex Motion

A cyclist starts with an initial velocity of 5 m/s and accelerates uniformly to 15 m/s in 10 seconds. Calculate the total displacement.

Solution:

The graph for this motion would be a straight line from (0 s, 5 m/s) to (10 s, 15 m/s). This forms a trapezium.

Area = ½ × (u + v) × t

Area = ½ × (5 m/s + 15 m/s) × 10 s

Area = ½ × (20 m/s) × 10 s

Area = 100 m

The total displacement of the cyclist is 100 m.