Equations of Motion

This lesson covers the SUVAT equations, applying them to problems, and free fall under gravity, as required by the Edexcel GCSE Combined Science specification (1SC0). These equations allow you to solve any problem involving uniform acceleration.

---

The Two Key Equations

At GCSE Combined Science level, you need to use two main equations of motion (sometimes called the SUVAT equations):

Equation 1

$v = u + at$v=u+at

This relates final velocity to initial velocity, acceleration and time.

Equation 2

$v^2 = u^2 + 2as$v2=u2+2as

This relates final velocity to initial velocity, acceleration and distance — useful when time is not given.

Variables

Symbol	Meaning	SI Unit
s	displacement (distance)	m
u	initial velocity	m/s
v	final velocity	m/s
a	acceleration	m/s²
t	time	s

Exam Tip: These two equations are given on the equation sheet in the exam. You must be able to select the right one and substitute values correctly. Choose the equation that contains the three quantities you know and the one you want to find.

---

Applying v = u + at

Worked Example 1

A car accelerates from 5 m/s to 25 m/s in 10 s. What is the acceleration?

Rearrange: $a = \frac{v - u}{t} = \frac{25 - 5}{10} = \frac{20}{10} = 2 \text{ m/s}^2$a=tv−u​=1025−5​=1020​=2 m/s2

Worked Example 2

A cyclist decelerates from 12 m/s at −1.5 m/s². How long until she stops?

When she stops, v = 0:

$0 = 12 + (-1.5) \times t$0=12+(−1.5)×t $1.5t = 12$1.5t=12 $t = \frac{12}{1.5} = 8 \text{ s}$t=1.512​=8 s

Worked Example 3

A train starts from rest with an acceleration of 0.8 m/s². What is its velocity after 30 s?

$v = 0 + 0.8 \times 30 = 24 \text{ m/s}$v=0+0.8×30=24 m/s

---

Applying v² = u² + 2as

Worked Example 4

A car accelerates from 10 m/s to 30 m/s with an acceleration of 4 m/s². What distance does it travel?

$v^2 = u^2 + 2as$v2=u2+2as $30^2 = 10^2 + 2 \times 4 \times s$302=102+2×4×s $900 = 100 + 8s$900=100+8s $8s = 800$8s=800 $s = 100 \text{ m}$s=100 m

Worked Example 5

A ball is thrown upward at 15 m/s. How high does it go? (g = 9.8 m/s², ignore air resistance.)

At the highest point, v = 0. Taking upward as positive, a = −9.8 m/s²:

$0^2 = 15^2 + 2 \times (-9.8) \times s$02=152+2×(−9.8)×s $0 = 225 - 19.6s$0=225−19.6s $19.6s = 225$19.6s=225 $s = \frac{225}{19.6} = 11.5 \text{ m}$s=19.6225​=11.5 m

Exam Tip: When using v² = u² + 2as, be careful with signs. If an object decelerates, a is negative. If an object falls, decide whether up or down is positive and be consistent.

---

Free Fall and Acceleration Due to Gravity

An object in free fall is acted on only by gravity (air resistance is ignored). All objects near the Earth's surface accelerate at the same rate:

$g = 9.8 \text{ m/s}^2$g=9.8 m/s2

This means:

• A falling object's speed increases by 9.8 m/s every second.
• The value of g does not depend on the mass of the object.

Worked Example 6

A stone is dropped from rest. What is its velocity after 3 s?

$v = u + at = 0 + 9.8 \times 3 = 29.4 \text{ m/s}$v=u+at=0+9.8×3=29.4 m/s

Worked Example 7

A stone is dropped from rest. How far does it fall in 3 s?

Using $v^2 = u^2 + 2as$v2=u2+2as:

$29.4^2 = 0^2 + 2 \times 9.8 \times s$29.42=02+2×9.8×s $864.36 = 19.6s$864.36=19.6s $s = 44.1 \text{ m}$s=44.1 m

Or using the area under the v–t graph: $s = \frac{1}{2} \times t \times v = \frac{1}{2} \times 3 \times 29.4 = 44.1 \text{ m}$s=21​×t×v=21​×3×29.4=44.1 m

---

Choosing the Right Equation

Known Quantities	Unknown	Equation to Use
u, a, t	v	$v = u + at$v=u+at
u, v, a	s	$v^2 = u^2 + 2as$v2=u2+2as
u, v, t	a	$a = (v - u) / t$a=(v−u)/t
u, a, s	v	$v^2 = u^2 + 2as$v2=u2+2as

Strategy for Solving Problems

1. List the known values (u, v, a, t, s).
2. Identify the unknown you need to find.
3. Choose the equation that contains those quantities.
4. Substitute and solve.
5. Check your answer is reasonable and include the correct unit.

Exam Tip: Always write out what you know (u = ..., v = ..., etc.) before selecting an equation. This systematic approach avoids errors and earns method marks even if you make an arithmetic slip.

---

Summary

• Two key equations of motion: $v = u + at$v=u+at and $v^2 = u^2 + 2as$v2=u2+2as.
• Use $v = u + at$v=u+at when time is known or required.
• Use $v^2 = u^2 + 2as$v2=u2+2as when time is not involved.
• Free fall: acceleration due to gravity g = 9.8 m/s², independent of mass.
• A dropped object starts with u = 0 and accelerates at 9.8 m/s² downward.
• Always list known values, choose the right equation, substitute carefully, and check units.
• Be consistent with signs when objects decelerate or move upward against gravity.

---

Extended Worked Examples

Example 8: Plane take-off

An aircraft accelerates uniformly along a 1200 m runway. It lifts off at 75 m/s. Find the acceleration if the aircraft starts from rest.

• Known: u = 0, v = 75 m/s, s = 1200 m.
• Use $v^2 = u^2 + 2as$v2=u2+2as.
• Rearrange: $a = (v^2 - u^2)/(2s) = (75^2 - 0)/2400 = 5625/2400 = 2.34$a=(v2−u2)/(2s)=(752−0)/2400=5625/2400=2.34 m/s² (3 s.f.).

Example 9: Mixed units

A train slows from 108 km/h to 36 km/h over 500 m. Find the deceleration in m/s².