Calculating Acceleration

Acceleration describes how quickly an object's velocity changes over time.

An object is accelerating whenever its velocity changes. This can happen when the object:

• speeds up
• slows down
• changes direction
• changes both speed and direction

Acceleration is therefore a vector quantity because it has both magnitude and direction.

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The Acceleration Equation

Average acceleration can be calculated using:

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

where:

• a = acceleration
• v_i = initial velocity
• v_f = final velocity
• t = time taken
• Δv = change in velocity

The symbol Δ

means change in.

Therefore:

Δv = v_f - v_i

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SI Units of Acceleration

The SI unit of velocity is: m/s

Acceleration measures the change in velocity per second.

Therefore:

\( \frac{m/s}{s} \)

which becomes:

m/s²

Acceleration is measured in metres per second squared.

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What Does m/s² Actually Mean?

Suppose a car has an acceleration of: 3 m/s²

This means its velocity changes by: 3 m/s every second.

If the car starts from rest:

After every second, another 3 m/s has been added to the velocity.

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Calculating Acceleration

Suppose a car increases its velocity from: 5 m/s to 17 m/s in 4 s

Use:

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

Substitute:

\( a = \frac{17 - 5}{4} = \frac{12}{4} = 3 m/s^2 \)

The car's velocity increases by an average of 3 m/s every second.

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A Useful Problem-Solving Method

Acceleration calculations can be organized into four steps.

Step 1 – Identify the information

Write down:

v_i = ?

v_f = ?

t = ?

Step 2 – Choose the equation

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

Step 3 – Substitute the values

Include the correct signs and units.

Step 4 – Calculate and interpret

Give the answer in: m/s²

and decide what the result tells you about the object's motion.

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Worked Example: Starting from Rest

A cyclist starts from rest and reaches: 12 m/s in 6 s

"Starts from rest" means:

v_i = 0

Therefore:

\( a = \frac{12 - 0}{6} = 2 m/s^2 \)

The cyclist's velocity increases by an average of 2 m/s every second.

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Worked Example: A Moving Object Speeds Up

A train is initially travelling at: 10 m/s

It reaches: 25 m/s

after: 5 s

Calculate:

\( a = \frac{25 - 10}{5} = \frac{15}{5} = 3 m/s^2 \)

Notice that the initial velocity was not zero.

Always check whether the object actually starts from rest.

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Deceleration

When an object slows down, its acceleration acts in the opposite direction to its velocity.

Suppose a car slows from: 20 m/s to 8 m/s in 4 s.

Calculate:

\( a = \frac{8 - 20}{4} = \frac{-12}{4} = -3 m/s^2 \)

If the positive direction is the direction the car is moving, the negative sign tells us the acceleration is acting in the opposite direction.

The car is decelerating.

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Accelerating or Decelerating?

A common shortcut is:

Speed increasing → accelerating

Speed decreasing → decelerating

But when working with signed velocities, we need to be more careful.

An object's speed increases when its velocity and acceleration point in the same direction.

Its speed decreases when velocity and acceleration point in opposite directions.

For example:

So negative acceleration does not always mean slowing down.

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Example with Negative Velocity

Suppose an object is moving left.

We define right as positive, so:

v_i = -4 m/s

Later:

v_f = - 10 m/s

Time: 3 s

Calculate:

\( a = \frac{-10 - (-4)}{3} = \frac{-6}{3} = -2 m/s^2 \)

The object's speed changed from:

4 m/s → 10 m/s

So it is actually speeding up in the negative direction.

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Zero Acceleration

Suppose a car travels at: 15 m/s

and five seconds later it is still travelling at: 15 m/s

in the same direction.

Then:

\( a = \frac{15 - 15}{5} = 0m/s^2 \)

The car has zero acceleration because its velocity is not changing.

This does not mean the car is stationary.

It means it is moving with constant velocity.

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Rearranging the Acceleration Equation

Sometimes acceleration is not the unknown quantity.

Starting with:

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

we can rearrange the equation to calculate other quantities.

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Finding Final Velocity

Start with:

\( a = \frac{ \Delta v }{ \Delta t } = \frac{v_f - v_i}{t} \)

Multiply both sides by t:

at = v_f - v_i

Add v_i:

v_f = v_i + at

This equation allows us to calculate final velocity when we know:

• initial velocity
• acceleration
• time

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Worked Example: Finding Final Velocity

A motorcycle travels initially at: 8 m/a

and accelerates at: 4 m/s²

for: 3 s

Use: v_f = v_i + at

Substitute:

v_f = = 8 + (4)(3) = 8 + 12 = 20 m/s

The motorcycle reaches a velocity of 20 m/s.

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Visualizing Constant Acceleration

The relationship between initial velocity, acceleration, time, and final velocity can also be seen on motion graphs.

For constant acceleration, equal time intervals produce equal changes in velocity.

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Finding Time

Starting again with:

\( a = \frac{v_f - v_i}{t} \)

Rearrange:

at = v_f - v_i

Then:

\( t = \frac{v_f - v_i}{a} \)

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Worked Example: Finding Time

A train increases its velocity from: 6 m/s to 24 m/s

with an acceleration of: 3 m/s²

Calculate the time.

\( t = \frac{24 - 6}{3} = \frac{18}{3} = 6 s \)

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Finding Initial Velocity

We can also rearrange:

v_f = v_i + at

to give:

v_i = v_f - at

Example

A car reaches: 30 m/s

after accelerating at: 2 m/s²

for: 5 s

Calculate its initial velocity.

v_i = 30 - (2)(5) = 30 - 10 = 20 m/s

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The Acceleration Equation Family

These equations are closely related:

Acceleration

\( a = \frac{v_f - v_i}{t} \)

Final Velocity

v_f = v_i + at

Initial Velocity

v_i = v_f - at

Time

\( t = \frac{v_f - v_i}{a} \)

Rather than memorizing four unrelated formulas, it is often better to understand how to rearrange the original acceleration equation.

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Worked Example: Braking

A bus is travelling at: 22 m/s

It brakes with an acceleration of: - 5.5 m/s²

How long does it take to stop?

At rest:

v_f = 0

Use:

\( t = \frac{v_f - v_i}{a} \)

Substitute:

\( t = \frac{0 - 22}{-5.5} = \frac{-22}{-5.5} = 4 s \)

The bus takes 4 seconds to stop.

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Worked Example: Free Fall

Ignoring air resistance, objects near Earth's surface have a downward acceleration of approximately:

g = 9.8 m/s²

Suppose a stone is dropped from rest.

Find its velocity after: 2.5 s

Take downward as positive.

v_i = 0

a = 9.8 m/s²

Use:

v_f = v_i + at = 0 + (9.8)(2.5) = 24.5 m/s downward

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Worked Example: More Challenging

A car travelling at: 72 km/h

brakes uniformly to rest in: 5 s

Find its acceleration.

The velocities should first be expressed in SI units.

Convert: 72 km/h

to m/s:

72 ÷ 3.6 = 20 m/s

Therefore:

v_i = 20 m/s

v_f = 0

Now:

\( a = \frac{0 - 20}{5} = -4 m/s^2 \)

The car's velocity decreases by 4 m/s each second.

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Converting km/h to m/s

Acceleration calculations commonly require velocities in: m/s

To convert:

km/h → m/s

Divide by 3.6

For example:

90 km/h ÷ 3.6 = = 25 m/s

m/s → km/h

Multiply by 3.6

For example:

(20 m/s)(3.6) = 72 km/h

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Acceleration from a Velocity-Time Graph

Acceleration can also be calculated from a velocity-time graph.

The slope of the graph represents acceleration:

\( slope = \frac{ \Delta v }{ \Delta t } \)

A graph sloping upward has positive acceleration.

A horizontal graph has: a = 0

A graph sloping downward has negative acceleration.

The steeper the graph, the greater the magnitude of the acceleration.

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Analysing Acceleration Data

Consider:

Every two seconds, velocity increases by: 4 m/s

Therefore:

\( a = \frac{4}{2} = 2 m/s^2 \)

Because the velocity increases by the same amount during each equal time interval, the object has constant acceleration.

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Quantitative Problem: Comparing Two Vehicles

Vehicle A increases from: 10 m/s → 25 m/s

in: 5 s

Vehicle B increases from:

5 m/s → 23 m/s

in: 4 s

Vehicle A

\( a_A = \frac{25 - 10}{5} = 3 m/s^2 \)

Vehicle B

\( a_B = \frac{23 - 5}{4} = 4.5 m/s^2 \)

Therefore:

Vehicle B has the greater acceleration.

Notice that comparing final velocities alone would not answer the question. We must consider both the change in velocity and the time taken.

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Checking Your Answer

After solving an acceleration problem, ask:

Are my units correct?

Acceleration should normally be: m/s²

Did I calculate the velocity change correctly?

Remember:

Δv = v_f - v_i

not: v_i - v_f

Does the sign make sense?

Check the chosen positive direction and whether the object's speed is increasing or decreasing.

Is the magnitude reasonable?

A calculation such as: 5000 m/s²

for an ordinary bicycle probably indicates an error.

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Common Mistakes

Forgetting the Initial Velocity

Incorrect: \( a = \frac{v_f}{t} \)

unless:

v_i = 0

Usually:

\( a = \frac{v_f - v_i}{t} \)

Using the Wrong Units

Velocity should normally be converted to: m/s

and time to: s

before calculating acceleration in SI units.

Assuming Negative Acceleration Always Means Slowing Down

Negative acceleration describes direction.

Whether the object speeds up or slows down depends on the directions of both velocity and acceleration.

Confusing Velocity with Acceleration

Velocity tells us how quickly position changes.

Acceleration tells us how quickly velocity changes.

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Did You Know?

The acceleration due to gravity near Earth's surface is approximately: 9.8 m/s²

This is often described using the symbol: g

Accelerations are sometimes compared using multiples of g.

For example: 2g ≈ 19.6 m/s²

Pilots, astronauts, roller-coaster riders, and racing drivers can temporarily experience accelerations described in terms of g-forces.

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Key Terms

Acceleration – Rate of change of velocity.

Initial velocity (v_i) – Velocity at the beginning of a time interval.

Final velocity (v_f) – Velocity at the end of a time interval.

Change in velocity (Δv) – Final velocity minus initial velocity.

Deceleration – A decrease in speed.

Constant acceleration – Acceleration that remains unchanged over time.

Metres per second squared (m/s²) – The SI unit of acceleration.

Free fall – Motion in which gravity is the only significant force acting on an object.

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Key Takeaways

• Acceleration measures how quickly velocity changes.
• The SI unit of acceleration is: m/s²
• Average acceleration is calculated using: \( a = \frac{v_f - v_i}{t} \)
• A positive or negative acceleration indicates the direction of acceleration, not automatically whether an object is speeding up or slowing down.
• An object speeds up when velocity and acceleration are in the same direction.
• An object slows down when velocity and acceleration are in opposite directions.
• Zero acceleration means constant velocity, not necessarily zero velocity.
• The acceleration equation can be rearranged to calculate velocity or time.
• Final velocity can be calculated using: v_f = v_i + at
• Time can be calculated using: \( t = \frac{v_f - v_i}{a} \)
• Initial velocity can be calculated using: v_i = v_f - at
• Velocities may need to be converted from km/h to m/s before calculating.
• The slope of a velocity-time graph represents acceleration.
• Quantitative acceleration problems require careful attention to values, signs, units, and direction.