5.6.1g Acceleration

Calculating average acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity, measured in metres per second squared ( $m/s2\text{m/s}^2$ ).

Average acceleration is $a=Δvta = \dfrac{\Delta v}{t}$ $a = \frac{Δ v}{t}$ .
1. $a$ is the acceleration in $m/s2\text{m/s}^2$ .
2. $Δv\Delta v$ is the change in velocity in $m/s\text{m/s}$ .
3. $t$ is the time taken in $s\text{s}$ .
The change in velocity is $velocity\Delta v = \text{final velocity} - \text{initial velocity}$ .
An acceleration of $m/s23\ \text{m/s}^2$ means the velocity changes by $m/s3\ \text{m/s}$ every second.

Example

A cyclist’s velocity increases from $m/s2\ \text{m/s}$ to $m/s8\ \text{m/s}$ in $s3\ \text{s}$ . Calculate the average acceleration.

Write the equation: $\dfrac{\Delta v}{t}$ .
Find the change in velocity: $m/s\Delta v = 8 - 2 = 6\ \text{m/s}$ .
Substitute the values: $\dfrac{6\ \text{m/s}}{3\ \text{s}}$ .
Calculate the acceleration: $2\ \text{m/s}^2$ .

Deceleration

Definition

Deceleration

Deceleration is the slowing down of an object.

When an object moving in the positive direction slows down, its change in velocity is negative, so its acceleration is negative.
A car slowing from $m/s15\ \text{m/s}$ to $m/s5\ \text{m/s}$ in $s2\ \text{s}$ has $m/s\Delta v = 5 - 15 = -10\ \text{m/s}$ , so $\dfrac{-10}{2} = -5\ \text{m/s}^2$ .
The magnitude of the deceleration is $m/s25\ \text{m/s}^2$ , because magnitude means size and is given as a positive value.

Common Mistake

Calculate the change in velocity as final velocity minus initial velocity, not the other way round.
Acceleration is measured in $m/s2\text{m/s}^2$ , not $m/s\text{m/s}$ .
A negative answer is not wrong; it shows the velocity decreasing in the chosen positive direction.
Deceleration is not a different quantity from acceleration; it is acceleration that slows an object down.

Estimating everyday accelerations

An estimate need not be exact but should be physically reasonable; everyday accelerations are often a few $m/s2\text{m/s}^2$ $m/s^{2}$ .
1. A cyclist going from $0$ to $m/s6\ \text{m/s}$ in $s3\ \text{s}$ accelerates at about $m/s22\ \text{m/s}^2$ .
2. A family car going from $0$ to $m/s20\ \text{m/s}$ in $s5\ \text{s}$ accelerates at about $m/s24\ \text{m/s}^2$ .
3. Hard braking gives a deceleration of magnitude several $m/s2\text{m/s}^2$ .
To check an estimate, ask whether the stated change in velocity could realistically happen in the stated time.

Acceleration from a velocity–time graph

Acceleration equals the gradient of a velocity–time graph: $a=change in velocitychange in timea = \dfrac{\text{change in velocity}}{\text{change in time}}$ $a = \frac{change in velocity}{change in time}$ , because velocity is on the vertical axis and time on the horizontal.
1. A straight line with a positive gradient shows constant positive acceleration.
2. A horizontal line shows constant velocity, so zero acceleration.
3. A downward-sloping line shows negative acceleration (deceleration if the velocity is positive).
A steeper line has a greater acceleration magnitude; a curved line shows changing acceleration.

Three velocity-time graphs showing constant velocity as a horizontal line, uniformly accelerated motion as a straight line with a positive gradient, and uniformly decelerated motion as a straight line with a negative gradient.

Example

Two points on a straight section of a velocity–time graph are $m/s)(2\ \text{s},\ 4\ \text{m/s})$ and $m/s)(7\ \text{s},\ 14\ \text{m/s})$ . Calculate the acceleration.

Write the gradient equation: $\dfrac{\text{change in velocity}}{\text{change in time}}$ .
Find the change in velocity: $10\ \text{m/s}$ .
Find the change in time: $5\ \text{s}$ .
Calculate the acceleration: $\dfrac{10}{5} = 2\ \text{m/s}^2$ .

Exam technique

Put time on the horizontal axis ( $s\text{s}$ ) and velocity on the vertical axis ( $m/s\text{m/s}$ ), with sensible, even scales.
Plot each measured time and velocity, draw the appropriate straight line or smooth curve, and use the gradient to find the acceleration.
For an acceleration calculation, state the equation, find $Δv\Delta v$ as final minus initial, substitute, and give $m/s2\text{m/s}^2$ .
For a graph, use two well-separated points and show the changes in velocity and time; if asked for the magnitude, give the positive size of the acceleration.

Self review

What is acceleration?
State the equation for average acceleration and its unit.
How do you calculate the change in velocity?
What does a horizontal line on a velocity–time graph mean?
How is acceleration found from a velocity–time graph?

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Definition

Area under a velocity–time graph

The area between a line on a velocity–time graph and the time axis represents the object's displacement.

The area between the line and the time axis represents the object's displacement, because $displacement=velocity×time\text{displacement}=\text{velocity}\times\text{time}$ .
The units confirm this, $ms×s=m\dfrac{\text{m}}{\text{s}}\times\text{s}=\text{m}$ .
If the velocity is constant, the area is a rectangle; if the velocity changes at a constant rate, the line is straight and the area is a triangle or trapezium.
For a journey with several stages, split the area into simple shapes and add their areas.
When the graph stays above the time axis, the magnitude of the displacement equals the distance travelled; an area below the axis represents displacement in the opposite direction.
To find displacement, add areas above the axis and subtract areas below it.
To find the distance travelled, add the magnitudes of all the areas, including those below the axis.

A velocity-time graph showing a straight line starting at initial velocity u and ending at final velocity v at time t. The area under the graph is shaded and divided into a rectangle (base t, height u) and a triangle (base t, height v-u), representing displacement.

Example

A cyclist accelerates uniformly from $m/s0\ \text{m/s}$ to $m/s12\ \text{m/s}$ in the first $s4\ \text{s}$ , then travels at $m/s12\ \text{m/s}$ for a further $s6\ \text{s}$ . Find the total distance travelled.

Area of the triangle during acceleration: $m\dfrac{1}{2}\times 4\ \text{s}\times 12\ \text{m/s}=24\ \text{m}$ .
Area of the rectangle at constant velocity: $m6\ \text{s}\times 12\ \text{m/s}=72\ \text{m}$ .
Add the areas: $m24\ \text{m}+72\ \text{m}=96\ \text{m}$ .
Answer: $m96\ \text{m}$ .

Estimating area by counting squares

For an irregular velocity–time graph, the area can be estimated by counting squares.
First find the area of one square, $square\text{area of one square}=\text{time per square}\times\text{velocity per square}$ .
Count the complete squares, combine the part-squares into whole ones, and multiply the total by the area of one square.
Always use the scales on both axes rather than assuming each square represents $m1\ \text{m}$ .

Uniform acceleration

Definition

Uniform acceleration

Uniform acceleration is acceleration that stays constant, so the velocity changes by equal amounts in equal time intervals.

On a velocity–time graph, uniform acceleration produces a straight line with a constant gradient.
For uniform acceleration, distance is linked to initial velocity, final velocity and acceleration by $v^{2} - u^{2} = 2 a s$ , where:
1. $v$ is the final velocity in $m/s\text{m/s}$ ;
2. $u$ is the initial velocity in $m/s\text{m/s}$ ;
3. $a$ is the acceleration in $m/s2\text{m/s}^2$ ;
4. $s$ is the distance in $m\text{m}$ .
This equation applies only when the acceleration is uniform.

Free fall near the Earth

An object in free fall accelerates under gravity; near the Earth's surface its acceleration is about $m/s2g=9.8\ \text{m/s}^2$ .
The value of $g$ is the acceleration, not the velocity; for a freely falling object, replace $a$ with $m/s29.8\ \text{m/s}^2$ .

Example

A stone is dropped from rest and falls freely through a distance of $m20\ \text{m}$ . Find its final velocity.

Use $v^2-u^2=2as$ with $m/su=0\ \text{m/s}$ , $m/s2a=9.8\ \text{m/s}^2$ and $ms=20\ \text{m}$ .
Substitute: $v2−02=2×9.8×20=392v^2-0^2=2\times 9.8\times 20=392$ .
Take the square root: $v=392v=\sqrt{392}$ .
Answer: $m/sv=19.8\ \text{m/s}$ downwards.

Common Mistake

Do not confuse the gradient and the area of a velocity–time graph: the gradient represents acceleration, while the area represents displacement.
When counting squares, always allow for the scale of both axes.
Areas below the time axis are negative for displacement but still count positively towards the total distance travelled.

Exam technique

For an area question, divide the region into rectangles, triangles or trapeziums, write the formula for each area, and include the unit $m\text{m}$ .
For an irregular graph, state the area represented by one square before you start counting.
When using $v^2-u^2=2as$ , write the equation first, substitute values with units, then remember to take the square root if you have found $v^2$ .

Choosing the right method

Use the area under the graph when a question asks for distance or displacement.
Use $v^2-u^2=2as$ when the acceleration is uniform and you know three of the four quantities.

Self review

What does the area under a velocity–time graph represent?
How do you find the distance travelled when part of the graph is below the time axis?
What graph shape shows uniform acceleration?
Which equation links $u$ , $v$ , $a$ and $s$ ?
What is the acceleration of a freely falling object near the Earth's surface?

5.6.1i Terminal velocity

Terminal velocity

Definition

Terminal velocity

Terminal velocity is the constant velocity reached by a falling object when the resistive force acting upwards is equal to the object’s weight acting downwards.

An object reaches its terminal velocity when the upward resistive force has grown until it equals the object’s weight.
A fluid is a liquid or a gas; when an object falls through a fluid, two main forces act on it.

The forces on a falling object

Definition

Resultant force

The resultant force is the single force that has the same effect as all the forces acting on an object combined.

Weight acts vertically downwards because of gravity.
Drag, such as air resistance, acts upwards, opposing the motion.
Whether the object speeds up, slows down or stays at a steady speed depends on the resultant force of these two forces.

How a falling object reaches terminal velocity

When first released from rest, the object’s weight acts downwards while the drag force is zero or very small, so the forces are unbalanced.
Weight is greater than drag, giving a resultant force downwards, so the object accelerates and its velocity increases.
As the object speeds up, the drag force increases while its weight stays constant, so the resultant force and the acceleration both decrease.
The velocity keeps increasing, but at a decreasing rate.
Eventually the drag force equals the weight; the resultant force is zero, so the object stops accelerating and falls at a constant terminal velocity.

Three free-body diagrams of a falling object: at release, only weight acts downwards; during acceleration, weight is greater than the upward drag force; at terminal velocity, weight and drag are equal and balanced.

Key Idea

At terminal velocity, weight equals drag.
The resultant force is zero.
The acceleration is zero.
The velocity is constant and is not zero.

Velocity–time graphs for terminal velocity

Plot time in seconds ( $s\text{s}$ ) on the horizontal axis and velocity in metres per second ( $m/s\text{m/s}$ ) on the vertical axis.
For an object dropped from rest, the line starts at the origin with a steep positive gradient, then curves and gradually becomes less steep, ending in a horizontal section.
The gradient represents acceleration: it is large at first because the resultant force is large, and it decreases as drag increases.
When terminal velocity is reached, the graph is horizontal; its gradient is zero, showing zero acceleration, and the height of this section is the terminal velocity.

Example

A falling object speeds up from $m/s0\ \text{m/s}$ to $m/s12\ \text{m/s}$ during the first $s3.0\ \text{s}$ , and later its velocity stays constant at $m/s18\ \text{m/s}$ . Find its average acceleration during the first $s3.0\ \text{s}$ and state its terminal velocity.

Use $\dfrac{v - u}{t}$ .
Substitute: $\dfrac{12 - 0}{3.0}$ .
Average acceleration: $m/s24.0\ \text{m/s}^2$ .
The terminal velocity is $m/s18\ \text{m/s}$ , the constant velocity shown by the horizontal section of the graph.

Interpreting the motion from the forces

The steep initial gradient corresponds to a large resultant force, because weight is much greater than drag.
As the curve flattens, drag is increasing and the resultant force is falling, so the acceleration is decreasing.
The horizontal section corresponds to drag equal to weight and a zero resultant force, so the object moves at a constant terminal velocity.

Common Mistake

A zero resultant force does not mean the object is stationary; at terminal velocity the object is still falling, but with zero acceleration because its velocity is constant.
Do not say the forces disappear; weight and drag both still act, but they are equal in size and opposite in direction.

Exam technique

Explain terminal velocity as a clear sequence that links the forces to the motion.
1. Start by stating that weight is greater than drag, giving a resultant force downwards, so the object accelerates.
2. Then explain that as the velocity increases the drag increases, so the resultant force and the acceleration decrease.
3. Finish by stating that when drag equals weight the resultant force is zero and the object moves at a constant terminal velocity.
For a velocity–time graph, refer to the gradient when discussing acceleration and identify the horizontal section as constant velocity.

Self review

What two main forces act on an object falling through a fluid?
Why does the drag force increase as the object speeds up?
What is the resultant force at terminal velocity?
What is the relationship between weight and drag at terminal velocity?
Why is the terminal-velocity section of a velocity–time graph horizontal?

Calculating average acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity, measured in metres per second squared ( $m/s2\text{m/s}^2$ ).

Average acceleration is $a=Δvta = \dfrac{\Delta v}{t}$ $a = \frac{Δ v}{t}$ .
1. $a$ is the acceleration in $m/s2\text{m/s}^2$ .
2. $Δv\Delta v$ is the change in velocity in $m/s\text{m/s}$ .
3. $t$ is the time taken in $s\text{s}$ .
The change in velocity is $velocity\Delta v = \text{final velocity} - \text{initial velocity}$ .
An acceleration of $m/s23\ \text{m/s}^2$ means the velocity changes by $m/s3\ \text{m/s}$ every second.

Example

A cyclist’s velocity increases from $m/s2\ \text{m/s}$ to $m/s8\ \text{m/s}$ in $s3\ \text{s}$ . Calculate the average acceleration.

Write the equation: $\dfrac{\Delta v}{t}$ .
Find the change in velocity: $m/s\Delta v = 8 - 2 = 6\ \text{m/s}$ .
Substitute the values: $\dfrac{6\ \text{m/s}}{3\ \text{s}}$ .
Calculate the acceleration: $2\ \text{m/s}^2$ .

Deceleration

Definition

Deceleration

Deceleration is the slowing down of an object.

When an object moving in the positive direction slows down, its change in velocity is negative, so its acceleration is negative.
A car slowing from $m/s15\ \text{m/s}$ to $m/s5\ \text{m/s}$ in $s2\ \text{s}$ has $m/s\Delta v = 5 - 15 = -10\ \text{m/s}$ , so $\dfrac{-10}{2} = -5\ \text{m/s}^2$ .
The magnitude of the deceleration is $m/s25\ \text{m/s}^2$ , because magnitude means size and is given as a positive value.

Common Mistake

Calculate the change in velocity as final velocity minus initial velocity, not the other way round.
Acceleration is measured in $m/s2\text{m/s}^2$ , not $m/s\text{m/s}$ .
A negative answer is not wrong; it shows the velocity decreasing in the chosen positive direction.
Deceleration is not a different quantity from acceleration; it is acceleration that slows an object down.

Estimating everyday accelerations

An estimate need not be exact but should be physically reasonable; everyday accelerations are often a few $m/s2\text{m/s}^2$ $m/s^{2}$ .
1. A cyclist going from $0$ to $m/s6\ \text{m/s}$ in $s3\ \text{s}$ accelerates at about $m/s22\ \text{m/s}^2$ .
2. A family car going from $0$ to $m/s20\ \text{m/s}$ in $s5\ \text{s}$ accelerates at about $m/s24\ \text{m/s}^2$ .
3. Hard braking gives a deceleration of magnitude several $m/s2\text{m/s}^2$ .
To check an estimate, ask whether the stated change in velocity could realistically happen in the stated time.

Acceleration from a velocity–time graph

Acceleration equals the gradient of a velocity–time graph: $a=change in velocitychange in timea = \dfrac{\text{change in velocity}}{\text{change in time}}$ $a = \frac{change in velocity}{change in time}$ , because velocity is on the vertical axis and time on the horizontal.
1. A straight line with a positive gradient shows constant positive acceleration.
2. A horizontal line shows constant velocity, so zero acceleration.
3. A downward-sloping line shows negative acceleration (deceleration if the velocity is positive).
A steeper line has a greater acceleration magnitude; a curved line shows changing acceleration.

Example

Two points on a straight section of a velocity–time graph are $m/s)(2\ \text{s},\ 4\ \text{m/s})$ and $m/s)(7\ \text{s},\ 14\ \text{m/s})$ . Calculate the acceleration.

Write the gradient equation: $\dfrac{\text{change in velocity}}{\text{change in time}}$ .
Find the change in velocity: $10\ \text{m/s}$ .
Find the change in time: $5\ \text{s}$ .
Calculate the acceleration: $\dfrac{10}{5} = 2\ \text{m/s}^2$ .

Exam technique

Put time on the horizontal axis ( $s\text{s}$ ) and velocity on the vertical axis ( $m/s\text{m/s}$ ), with sensible, even scales.
Plot each measured time and velocity, draw the appropriate straight line or smooth curve, and use the gradient to find the acceleration.
For an acceleration calculation, state the equation, find $Δv\Delta v$ as final minus initial, substitute, and give $m/s2\text{m/s}^2$ .
For a graph, use two well-separated points and show the changes in velocity and time; if asked for the magnitude, give the positive size of the acceleration.

Self review

What is acceleration?
State the equation for average acceleration and its unit.
How do you calculate the change in velocity?
What does a horizontal line on a velocity–time graph mean?
How is acceleration found from a velocity–time graph?

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Definition

Area under a velocity–time graph

The area between a line on a velocity–time graph and the time axis represents the object's displacement.

The area between the line and the time axis represents the object's displacement, because $displacement=velocity×time\text{displacement}=\text{velocity}\times\text{time}$ .
The units confirm this, $ms×s=m\dfrac{\text{m}}{\text{s}}\times\text{s}=\text{m}$ .
If the velocity is constant, the area is a rectangle; if the velocity changes at a constant rate, the line is straight and the area is a triangle or trapezium.
For a journey with several stages, split the area into simple shapes and add their areas.
When the graph stays above the time axis, the magnitude of the displacement equals the distance travelled; an area below the axis represents displacement in the opposite direction.
To find displacement, add areas above the axis and subtract areas below it.
To find the distance travelled, add the magnitudes of all the areas, including those below the axis.

Example

Area of the triangle during acceleration: $m\dfrac{1}{2}\times 4\ \text{s}\times 12\ \text{m/s}=24\ \text{m}$ .
Area of the rectangle at constant velocity: $m6\ \text{s}\times 12\ \text{m/s}=72\ \text{m}$ .
Add the areas: $m24\ \text{m}+72\ \text{m}=96\ \text{m}$ .
Answer: $m96\ \text{m}$ .

Estimating area by counting squares

For an irregular velocity–time graph, the area can be estimated by counting squares.
First find the area of one square, $square\text{area of one square}=\text{time per square}\times\text{velocity per square}$ .
Count the complete squares, combine the part-squares into whole ones, and multiply the total by the area of one square.
Always use the scales on both axes rather than assuming each square represents $m1\ \text{m}$ .

Uniform acceleration

Definition

Uniform acceleration

Uniform acceleration is acceleration that stays constant, so the velocity changes by equal amounts in equal time intervals.

On a velocity–time graph, uniform acceleration produces a straight line with a constant gradient.
For uniform acceleration, distance is linked to initial velocity, final velocity and acceleration by $v^{2} - u^{2} = 2 a s$ , where:
1. $v$ is the final velocity in $m/s\text{m/s}$ ;
2. $u$ is the initial velocity in $m/s\text{m/s}$ ;
3. $a$ is the acceleration in $m/s2\text{m/s}^2$ ;
4. $s$ is the distance in $m\text{m}$ .
This equation applies only when the acceleration is uniform.

Free fall near the Earth

An object in free fall accelerates under gravity; near the Earth's surface its acceleration is about $m/s2g=9.8\ \text{m/s}^2$ .
The value of $g$ is the acceleration, not the velocity; for a freely falling object, replace $a$ with $m/s29.8\ \text{m/s}^2$ .

Example

A stone is dropped from rest and falls freely through a distance of $m20\ \text{m}$ . Find its final velocity.

Use $v^2-u^2=2as$ with $m/su=0\ \text{m/s}$ , $m/s2a=9.8\ \text{m/s}^2$ and $ms=20\ \text{m}$ .
Substitute: $v2−02=2×9.8×20=392v^2-0^2=2\times 9.8\times 20=392$ .
Take the square root: $v=392v=\sqrt{392}$ .
Answer: $m/sv=19.8\ \text{m/s}$ downwards.

Common Mistake

Do not confuse the gradient and the area of a velocity–time graph: the gradient represents acceleration, while the area represents displacement.
When counting squares, always allow for the scale of both axes.
Areas below the time axis are negative for displacement but still count positively towards the total distance travelled.

Exam technique

For an area question, divide the region into rectangles, triangles or trapeziums, write the formula for each area, and include the unit $m\text{m}$ .
For an irregular graph, state the area represented by one square before you start counting.
When using $v^2-u^2=2as$ , write the equation first, substitute values with units, then remember to take the square root if you have found $v^2$ .

Choosing the right method

Use the area under the graph when a question asks for distance or displacement.
Use $v^2-u^2=2as$ when the acceleration is uniform and you know three of the four quantities.

Self review

What does the area under a velocity–time graph represent?
How do you find the distance travelled when part of the graph is below the time axis?
What graph shape shows uniform acceleration?
Which equation links $u$ , $v$ , $a$ and $s$ ?
What is the acceleration of a freely falling object near the Earth's surface?

5.6.1i Terminal velocity

Terminal velocity

Definition

Terminal velocity

Terminal velocity is the constant velocity reached by a falling object when the resistive force acting upwards is equal to the object’s weight acting downwards.

An object reaches its terminal velocity when the upward resistive force has grown until it equals the object’s weight.
A fluid is a liquid or a gas; when an object falls through a fluid, two main forces act on it.

The forces on a falling object

Definition

Resultant force

The resultant force is the single force that has the same effect as all the forces acting on an object combined.

Weight acts vertically downwards because of gravity.
Drag, such as air resistance, acts upwards, opposing the motion.
Whether the object speeds up, slows down or stays at a steady speed depends on the resultant force of these two forces.

How a falling object reaches terminal velocity

When first released from rest, the object’s weight acts downwards while the drag force is zero or very small, so the forces are unbalanced.
Weight is greater than drag, giving a resultant force downwards, so the object accelerates and its velocity increases.
As the object speeds up, the drag force increases while its weight stays constant, so the resultant force and the acceleration both decrease.
The velocity keeps increasing, but at a decreasing rate.
Eventually the drag force equals the weight; the resultant force is zero, so the object stops accelerating and falls at a constant terminal velocity.

Key Idea

At terminal velocity, weight equals drag.
The resultant force is zero.
The acceleration is zero.
The velocity is constant and is not zero.

Velocity–time graphs for terminal velocity

Plot time in seconds ( $s\text{s}$ ) on the horizontal axis and velocity in metres per second ( $m/s\text{m/s}$ ) on the vertical axis.
For an object dropped from rest, the line starts at the origin with a steep positive gradient, then curves and gradually becomes less steep, ending in a horizontal section.
The gradient represents acceleration: it is large at first because the resultant force is large, and it decreases as drag increases.
When terminal velocity is reached, the graph is horizontal; its gradient is zero, showing zero acceleration, and the height of this section is the terminal velocity.

Example

Use $\dfrac{v - u}{t}$ .
Substitute: $\dfrac{12 - 0}{3.0}$ .
Average acceleration: $m/s24.0\ \text{m/s}^2$ .
The terminal velocity is $m/s18\ \text{m/s}$ , the constant velocity shown by the horizontal section of the graph.

Interpreting the motion from the forces

The steep initial gradient corresponds to a large resultant force, because weight is much greater than drag.
As the curve flattens, drag is increasing and the resultant force is falling, so the acceleration is decreasing.
The horizontal section corresponds to drag equal to weight and a zero resultant force, so the object moves at a constant terminal velocity.

Common Mistake

A zero resultant force does not mean the object is stationary; at terminal velocity the object is still falling, but with zero acceleration because its velocity is constant.
Do not say the forces disappear; weight and drag both still act, but they are equal in size and opposite in direction.

Exam technique

Explain terminal velocity as a clear sequence that links the forces to the motion.
1. Start by stating that weight is greater than drag, giving a resultant force downwards, so the object accelerates.
2. Then explain that as the velocity increases the drag increases, so the resultant force and the acceleration decrease.
3. Finish by stating that when drag equals weight the resultant force is zero and the object moves at a constant terminal velocity.
For a velocity–time graph, refer to the gradient when discussing acceleration and identify the horizontal section as constant velocity.

Self review

What two main forces act on an object falling through a fluid?
Why does the drag force increase as the object speeds up?
What is the resultant force at terminal velocity?
What is the relationship between weight and drag at terminal velocity?
Why is the terminal-velocity section of a velocity–time graph horizontal?

Calculating average acceleration

Deceleration

Estimating everyday accelerations

Acceleration from a velocity–time graph

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Estimating area by counting squares

Uniform acceleration

Free fall near the Earth

Choosing the right method

5.6.1i Terminal velocity

Terminal velocity

The forces on a falling object

How a falling object reaches terminal velocity

Velocity–time graphs for terminal velocity

Interpreting the motion from the forces

Review 5.6.1g Acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.1g Acceleration

Calculating average acceleration

Deceleration

Estimating everyday accelerations

Acceleration from a velocity–time graph

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Estimating area by counting squares

Uniform acceleration

Free fall near the Earth

Choosing the right method

5.6.1i Terminal velocity

Terminal velocity

The forces on a falling object

How a falling object reaches terminal velocity

Velocity–time graphs for terminal velocity

Interpreting the motion from the forces

Review 5.6.1g Acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.1g Acceleration

Calculating average acceleration

Deceleration

Estimating everyday accelerations

Acceleration from a velocity–time graph

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Estimating area by counting squares

Uniform acceleration

Free fall near the Earth

Choosing the right method

5.6.1i Terminal velocity

Terminal velocity

The forces on a falling object

How a falling object reaches terminal velocity

Velocity–time graphs for terminal velocity

Interpreting the motion from the forces

Review 5.6.1g Acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.1g Acceleration Revision Guide

5.6.1g Acceleration

Calculating average acceleration

Deceleration

Estimating everyday accelerations

Acceleration from a velocity–time graph

5.6.1h Distance from a velocity–time graph and uniform acceleration (HT)

Area under a velocity–time graph

Estimating area by counting squares

Uniform acceleration

Free fall near the Earth

Choosing the right method

5.6.1i Terminal velocity

Terminal velocity

The forces on a falling object

How a falling object reaches terminal velocity

Velocity–time graphs for terminal velocity

Interpreting the motion from the forces

Review 5.6.1g Acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.6.1g Acceleration Revision Guide