Why Rearrange Equations?

Physics equations describe relationships between physical quantities. Often, the equation you know is not arranged for the variable you need to calculate.

For example:

v = d/t

This equation is arranged to calculate velocity.

But what if you know velocity and time and need to calculate distance?

The equation must be rearranged:

d = vt

Or, if you need time:

t = d/v

The relationship between the quantities has not changed. We have simply changed which variable is isolated.

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What Does "Isolate a Variable" Mean?

To isolate a variable means to get that variable by itself on one side of the equation.

For example:

F = ma

Here, F is already isolated.

To isolate m:

m = F/a

To isolate a:

a = F/m

The goal of rearranging is usually:

wanted variable = everything else

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The Golden Rule of Algebra

The most important rule is:

Whatever you do to one side of an equation, you must do to the other side.

Think of an equation as a balanced scale.

If:

x + 5 = 12

subtract 5 from both sides:

x + 5 − 5 = 12 − 5

Therefore:

x = 7

The equation remains balanced.

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Inverse Operations

Rearranging equations depends on inverse operations.

Inverse operations undo each other.

For example, if a variable is multiplied by 4:

4x = 20

divide both sides by 4:

x = 20/4

x = 5

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Undo Operations in Reverse Order

Suppose:

3x + 5 = 20

The operations performed on x are:

x → multiply by 3 → add 5

To isolate x, undo them in reverse order:

subtract 5 → divide by 3

Start:

3x + 5 = 20

Subtract 5:

3x = 15

Divide by 3:

x = 5

This idea becomes extremely useful in multi-step physics equations.

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Rearranging Multiplication Equations

Consider Newton's second law:

F = ma

Suppose we want m.

Since m is multiplied by a, divide both sides by a:

F/a = ma/a

The a terms cancel:

m = F/a

Similarly, to find acceleration:

a = F/m

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Worked Example: Force

A 6.0 kg object accelerates at 4.0 m/s².

Use:

F = ma

Substitute:

F = 6.0 × 4.0

F = 24 N

No rearrangement was necessary because F was already isolated.

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Worked Example: Finding Mass

A force of 45 N produces an acceleration of 3.0 m/s².

Start with:

F = ma

We need m.

Divide by a:

m = F/a

Substitute:

m = 45/3.0

m = 15 kg

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Worked Example: Finding Acceleration

A resultant force of 120 N acts on a 30 kg object.

Start with:

F = ma

Rearrange:

a = F/m

Substitute:

a = 120/30

a = 4.0 m/s²

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Rearranging Division Equations

Consider:

v = d/t

Suppose we want distance.

Multiply both sides by t:

vt = d

Usually we write the wanted variable first:

d = vt

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Finding the Denominator

Suppose we want time from:

v = d/t

Multiply by t:

vt = d

Now divide by v:

t = d/v

So the three useful forms are:

v = d/t

d = vt

t = d/v

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Worked Example: Finding Time

A cyclist travels 1500 m at an average speed of 6.0 m/s.

Start with:

v = d/t

Rearrange:

t = d/v

Substitute:

t = 1500/6.0

t = 250 s

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Rearranging Density

Density is:

ρ = m/V

To find mass:

m = ρV

To find volume:

V = m/ρ

So:

ρ = m/V

m = ρV

V = m/ρ

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Worked Example: Density

An object has a density of 2700 kg/m³ and a volume of 0.0020 m³.

Find its mass.

Start with:

ρ = m/V

Rearrange:

m = ρV

Substitute:

m = 2700 × 0.0020

m = 5.4 kg

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Rearranging Pressure

Pressure is:

p = F/A

Possible forms are:

p = F/A

F = pA

A = F/p

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Worked Example: Area

A force of 500 N produces a pressure of 2500 Pa.

Find the area.

Start with:

p = F/A

Rearrange:

A = F/p

Substitute:

A = 500/2500

A = 0.20 m²

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Rearranging Weight

Weight is calculated using:

W = mg

To find mass:

m = W/g

To find gravitational field strength:

g = W/m

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Worked Example: Mass from Weight

An object weighs 196 N where:

g = 9.8 N/kg

Use:

W = mg

Rearrange:

m = W/g

Substitute:

m = 196/9.8

m = 20 kg

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Rearrange Before Substituting

A very useful habit is:

equation → rearrange → substitute → calculate

For example, suppose:

F = ma

and you need mass.

Do this:

F = ma

m = F/a

Then insert the numbers.

This is usually clearer than inserting numbers first and then trying to manipulate them.

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Rearranging Equations with Addition

Consider:

v = u + at

Suppose we want u.

Subtract at from both sides:

v − at = u

Therefore:

u = v − at

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Finding Acceleration

Start with:

v = u + at

We want a.

Subtract u:

v − u = at

Divide by t:

a = (v − u)/t

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Finding Time

Start with:

v = u + at

Subtract u:

v − u = at

Divide by a:

t = (v − u)/a

So the equation can be rearranged into several useful forms:

v = u + at

u = v − at

a = (v − u)/t

t = (v − u)/a

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Worked Example: Acceleration

A car changes velocity from 8 m/s to 26 m/s in 6.0 s.

Use:

v = u + at

Rearrange:

a = (v − u)/t

Substitute:

a = (26 − 8)/6.0

a = 18/6.0

a = 3.0 m/s²

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Brackets Matter

Consider:

a = (v − u)/t

The entire difference:

v − u

must be divided by t.

If:

v = 30 m/s

u = 10 m/s

t = 5 s

then:

a = (30 − 10)/5

a = 20/5

a = 4 m/s²

Writing:

30 − 10/5

would give a completely different answer because division occurs before subtraction.

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Rearranging Equations Containing Squares

Consider kinetic energy:

Eₖ = ½mv²

Suppose we want mass.

Multiply by 2:

2Eₖ = mv²

Divide by v²:

m = 2Eₖ/v²

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Finding Velocity from Kinetic Energy

Start with:

Eₖ = ½mv²

Multiply by 2:

2Eₖ = mv²

Divide by m:

2Eₖ/m = v²

Now take the square root:

v = √(2Eₖ/m)

This is a multi-step rearrangement.

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Worked Example: Velocity from Kinetic Energy

An object has:

Eₖ = 400 J

m = 8.0 kg

Use:

v = √(2Eₖ/m)

Substitute:

v = √((2 × 400)/8.0)

v = √100

v = 10 m/s

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Squaring and Square Roots Are Inverse Operations

If:

x² = 49

then:

x = √49

For a physical quantity such as speed, we normally use the physically appropriate positive value:

x = 7

This principle appears frequently in physics equations involving:

v²

r²

t²

and other squared quantities.

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Rearranging Gravitational Potential Energy

Consider:

Eₚ = mgh

To find mass:

m = Eₚ/(gh)

To find gravitational field strength:

g = Eₚ/(mh)

To find height:

h = Eₚ/(mg)

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Worked Example: Finding Height

An object has:

Eₚ = 980 J

m = 20 kg

g = 9.8 N/kg

Start with:

Eₚ = mgh

Rearrange:

h = Eₚ/(mg)

Substitute:

h = 980/(20 × 9.8)

h = 980/196

h = 5.0 m

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Rearranging Work Done

For a constant force acting in the direction of motion:

W = Fd

To find force:

F = W/d

To find distance:

d = W/F

Example

If 600 J of work is done by a 150 N force:

d = W/F

d = 600/150

d = 4.0 m

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Rearranging Power

Power is:

P = E/t

To find energy:

E = Pt

To find time:

t = E/P

Example

A 500 W device transfers 15,000 J of energy.

Find the time.

t = E/P

t = 15,000/500

t = 30 s

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Rearranging Momentum

Momentum is:

p = mv

To find mass:

m = p/v

To find velocity:

v = p/m

Example

An object has momentum:

p = 48 kg·m/s

and mass:

m = 6.0 kg

Therefore:

v = p/m

v = 48/6.0

v = 8.0 m/s

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Rearranging Ohm's Law

Ohm's law is:

V = IR

To find current:

I = V/R

To find resistance:

R = V/I

The same algebraic rules work regardless of the physics topic.

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Equations with Fractions on Both Sides

Consider:

ρ₁ = m₁/V₁

If you need V₁:

ρ₁V₁ = m₁

Then:

V₁ = m₁/ρ₁

Work systematically rather than trying to move symbols mentally.

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Equations with a Constant

Consider:

F = kx

To find x:

x = F/k

To find k:

k = F/x

The same principle applies to Hooke's law:

F = kx

where k is the spring constant and x is extension.

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Multi-Step Rearranging

Consider:

s = ut + ½at²

Suppose we need acceleration.

Start:

s = ut + ½at²

Subtract ut:

s − ut = ½at²

Multiply by 2:

2(s − ut) = at²

Divide by t²:

a = 2(s − ut)/t²

This is a good example of why rearranging should be done one operation at a time.

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Worked Multi-Step Example

A vehicle travels 100 m in 5.0 s with an initial velocity of 10 m/s.

Find its acceleration.

Use:

s = ut + ½at²

Rearrange:

a = 2(s − ut)/t²

Substitute:

a = 2(100 − (10 × 5.0))/(5.0)²

Calculate inside the brackets:

a = 2(100 − 50)/25

a = 100/25

a = 4.0 m/s²

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Rearranging an Equation with the Variable in a Denominator

Suppose:

P = E/t

and we need t.

Start:

P = E/t

Multiply by t:

Pt = E

Divide by P:

t = E/P

A useful rule is:

If the wanted variable is in the denominator, first multiply to bring it out of the denominator.

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Rearranging an Inverse-Square Equation

Consider:

F = k/r²

Suppose we need r.

Multiply by r²:

Fr² = k

Divide by F:

r² = k/F

Take the square root:

r = √(k/F)

This type of rearrangement occurs in gravitational, electric, and other inverse-square relationships.

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A More Complex Example: Gravitation

Newton's law of gravitation is:

F = Gm₁m₂/r²

Suppose we want r.

Multiply by r²:

Fr² = Gm₁m₂

Divide by F:

r² = Gm₁m₂/F

Take the square root:

r = √(Gm₁m₂/F)

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Rearranging for One of the Masses

Start with:

F = Gm₁m₂/r²

Suppose we need m₂.

Multiply by r²:

Fr² = Gm₁m₂

Divide by Gm₁:

m₂ = Fr²/(Gm₁)

This demonstrates how even complicated physics equations can be rearranged using the same basic algebraic rules.

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Substituting Known Values

Once the equation is rearranged, substitute the known values.

For example:

a = F/m

If:

F = 75 N

m = 15 kg

then:

a = 75/15

a = 5.0 m/s²

Include units when recording the known quantities and always include the correct unit in the final answer.

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Use Brackets During Substitution

Suppose:

m = 2Eₖ/v²

and:

Eₖ = 900 J

v = 15 m/s

Write:

m = (2 × 900)/(15)²

rather than entering a long expression without brackets.

Calculate:

m = 1800/225

m = 8.0 kg

Brackets reduce calculator-entry errors.

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Check Using the Original Equation

One of the best ways to check a rearrangement is to substitute the rearranged expression back into the original equation.

Suppose:

F = ma

and you rearranged to:

m = F/a

Substitute this into the original equation:

F = (F/a)a

The a terms cancel:

F = F

So the rearrangement is correct.

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Check Using Units

Units can also test a rearrangement.

Suppose:

m = F/a

Force has units:

N = kg·m/s²

Acceleration has units:

m/s²

Therefore:

m = (kg·m/s²)/(m/s²)

The m/s² terms cancel:

m = kg

The result has the correct unit for mass.

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Checking a Speed Rearrangement

Start with:

v = d/t

Suppose:

t = d/v

Check the units:

t = m/(m/s)

Dividing by m/s is equivalent to multiplying by s/m:

t = m × s/m

The metres cancel:

t = s

The result has the correct unit for time.

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Checking a Kinetic Energy Rearrangement

Suppose:

v = √(2Eₖ/m)

Energy has units:

J = kg·m²/s²

Therefore:

Eₖ/m = (kg·m²/s²)/kg

= m²/s²

Taking the square root gives:

m/s

That is the correct unit for velocity.

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Equation Triangles

For simple equations, some students use equation triangles.

For:

v = d/t

a triangle might place:

d

above:

v t

This can help remember:

d = vt

v = d/t

t = d/v

However, equation triangles have limitations.

They work well for simple multiplication and division relationships but are much less useful for equations involving:

• Addition or subtraction.
• Squares.
• Square roots.
• Several terms.
• Brackets.
• Powers.
• Variables appearing more than once.

Learning algebraic rearrangement is therefore much more powerful.

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A General Rearranging Strategy

When rearranging a physics equation:

Identify the variable you need.

↓

Find what operations are being performed on it.

↓

Undo those operations using inverse operations.

↓

Perform the same operation on both sides.

↓

Continue until the variable is isolated.

↓

Simplify the equation.

↓

Check the rearrangement.

↓

Substitute the known values.

↓

Calculate and include units.

This method works for equations across physics.

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Worked Example: A Full Problem

A car increases its velocity from 12 m/s to 30 m/s with an acceleration of 3.0 m/s².

How long does the acceleration take?

Choose the equation

v = u + at

Identify the wanted variable

We need:

t

Rearrange

Subtract u:

v − u = at

Divide by a:

t = (v − u)/a

Substitute

t = (30 − 12)/3.0

Calculate

t = 18/3.0

t = 6.0 s

Check

Use the original equation:

v = u + at

v = 12 + (3.0 × 6.0)

v = 30 m/s

The answer is consistent.

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Worked Example: Force and Area

A pressure of 25,000 Pa is produced by a force acting over an area of 0.040 m².

Find the force.

Start:

p = F/A

We need F.

Multiply by A:

F = pA

Substitute:

F = 25,000 × 0.040

F = 1000 N

Check units:

Pa × m²

Since:

Pa = N/m²

then:

N/m² × m² = N

The units confirm that the rearrangement makes sense.

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Worked Example: Finding Volume

A material has:

mass = 540 kg

density = 2700 kg/m³

Find its volume.

Start:

ρ = m/V

Rearrange:

V = m/ρ

Substitute:

V = 540/2700

V = 0.20 m³

Check the units:

kg ÷ kg/m³

= m³

Correct.

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Worked Example: Finding a Spring Constant

A spring extends 0.080 m when a force of 20 N is applied.

Use:

F = kx

We need k.

Rearrange:

k = F/x

Substitute:

k = 20/0.080

k = 250 N/m

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Worked Example: Finding Distance from Kinetic Energy

Suppose a constant braking force of 4000 N removes 120,000 J of kinetic energy from a car.

Using the magnitude of the work-energy relationship:

W = Fd

Find the braking distance.

Rearrange:

d = W/F

Substitute:

d = 120,000/4000

d = 30 m

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Common Rearrangements Worth Recognizing

Recognizing common patterns can make rearranging faster, but understanding the algebra is more important than memorizing the table.

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Common Mistakes

Moving a Term Without Changing the Operation

Students sometimes think terms simply "move across the equals sign."

For example:

F = ma

It is better to think:

divide both sides by a

rather than:

move a underneath F

Understanding the operation prevents mistakes in more complicated equations.

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Performing an Operation on Only One Side

Incorrect:

3x + 5 = 20

3x = 20

Correct:

3x + 5 = 20

Subtract 5 from both sides:

3x = 15

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Forgetting Brackets

Suppose:

a = (v − u)/t

Writing:

a = v − u/t

changes the equation.

Use brackets whenever several terms must be treated together.

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Forgetting to Square a Quantity

If:

Eₖ = ½mv²

then velocity is squared.

If:

v = 10 m/s

then:

v² = 100 m²/s²

not 20.

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Forgetting the Square Root

From:

v² = 100

you cannot conclude:

v = 100

You need:

v = √100

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Substituting Too Early

Putting numbers into a complicated equation before rearranging can make the algebra harder to follow.

A clearer method is:

formula → rearrange → substitute → calculate

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Ignoring Units

A numerical answer should include an appropriate unit.

If you calculate acceleration:

a = 5.0

the answer is incomplete.

Write:

a = 5.0 m/s²

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Using the Wrong Variable

Physics sometimes uses the same letter for different quantities in different contexts.

For example, p may represent pressure or momentum depending on the equation.

Always identify what each symbol represents.

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Check Your Understanding

1. Rearrange F = ma to make m the subject.

2. Rearrange F = ma to make a the subject.

3. Rearrange v = d/t to make d the subject.

4. Rearrange v = d/t to make t the subject.

5. Rearrange ρ = m/V to make V the subject.

6. Rearrange p = F/A to make F the subject.

7. Rearrange W = mg to make g the subject.

8. Rearrange P = E/t to make t the subject.

9. Rearrange V = IR to make R the subject.

10. Rearrange F = kx to make x the subject.

11. Rearrange v = u + at to make u the subject.

12. Rearrange v = u + at to make a the subject.

13. Rearrange v = u + at to make t the subject.

14. Rearrange Eₚ = mgh to make h the subject.

15. Rearrange Eₖ = ½mv² to make m the subject.

16. Rearrange Eₖ = ½mv² to make v the subject.

17. Rearrange F = k/r² to make r the subject.

18. Rearrange s = ut + ½at² to make a the subject.

19. A 12 kg object experiences a resultant force of 72 N. Calculate its acceleration.

20. An object travels 450 m at 15 m/s. Calculate the travel time.

21. A material has a mass of 780 kg and density of 2600 kg/m³. Calculate its volume.

22. A 6.0 kg object has 300 J of kinetic energy. Calculate its speed.

23. A vehicle changes velocity from 5.0 m/s to 29 m/s in 8.0 s. Calculate its acceleration.

24. A 15 kg object has 1470 J of gravitational potential energy where g = 9.8 N/kg. Calculate its height.

25. Explain two methods that can be used to check whether an equation has been rearranged correctly.

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Key Terms

• Equation – mathematical statement showing that two expressions are equal.
• Variable – symbol representing a quantity that can have different values.
• Subject – variable isolated on one side of an equation.
• Rearrange – change the form of an equation while preserving the relationship between its variables.
• Isolate – get a variable by itself on one side of an equation.
• Inverse operation – operation that reverses another operation.
• Substitute – replace a variable with a known value.
• Square – multiply a quantity by itself.
• Square root – inverse operation of squaring.
• Formula – equation describing a mathematical or physical relationship.

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Key Takeaways

• Rearranging an equation allows you to calculate different variables using the same physical relationship.
• The equation must remain balanced.
• Whatever operation is performed on one side must also be performed on the other.
• Use inverse operations to isolate the wanted variable.
• Addition is undone by subtraction.
• Multiplication is undone by division.
• Squaring is undone by taking a square root.
• Work one step at a time with complicated equations.
• Brackets are important when several terms must be treated together.
• Rearrange the equation before substituting numerical values whenever practical.
• Include units when substituting and in the final answer.
• Check rearrangements by substituting them back into the original equation.
• Dimensional analysis provides another powerful check.
• Multi-step equations use exactly the same algebraic principles as simple equations.

A reliable physics workflow is:

Identify the unknown → choose the equation → rearrange → check → substitute → calculate → check the answer.