Equations of motion in Physics: how to choose the right one (with worked examples)

The equations of motion are among the most tested ideas in senior secondary Physics. They appear in JAMB objective questions, in WAEC and NECO theory papers, and inside many longer problems on forces, momentum and energy. The good news: there are only three equations, and a simple method tells you which one to use every time.

The five quantities

Every uniformly accelerated motion problem involves five quantities:

The three equations

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as

You may also see a fourth, which is useful when acceleration isn't given:

4. s = ½(u + v)t

These equations only apply when acceleration is constant (uniform). For JAMB, WAEC and NECO questions on this topic, that is almost always the case.

The method: list, identify, choose

1. List what you know. Write down the values of u, v, a, t and s that the question gives you, including hidden ones.
2. Identify what you need.
3. Choose the equation that contains your unknown and the quantities you know, and leaves out the one you don't know and don't need.

So if a question gives you nothing about time and doesn't ask about it, use v² = u² + 2as.

Hidden information to look out for

Questions often hide values in words:

• "Starts from rest" → u = 0
• "Comes to rest" / "stops" → v = 0
• "Dropped" / "released" → u = 0
• "Free fall" → a = g ≈ 10 m/s² (use the value given in the question; some use 9.8 m/s²)
• "Decelerates" / "retards" → a is negative
• "Thrown vertically upwards… at its highest point" → v = 0

Worked example 1: A bus accelerating

A bus starts from rest and accelerates uniformly at 2 m/s² for 8 s. Find (a) its final velocity and (b) the distance covered.

List: u = 0 (starts from rest), a = 2 m/s², t = 8 s.

(a) Need v. Use v = u + at: v = 0 + 2 × 8 = 16 m/s

(b) Need s. Use s = ut + ½at²: s = 0 × 8 + ½ × 2 × 8² = 0 + 64 = 64 m

Worked example 2: Braking

A car travelling at 20 m/s brakes uniformly and stops in 50 m. Find its deceleration.

List: u = 20 m/s, v = 0 (stops), s = 50 m. Time isn't given or asked for.

Use v² = u² + 2as: 0 = 20² + 2 × a × 50 0 = 400 + 100a a = −4 m/s²

The negative sign shows deceleration. The deceleration is 4 m/s².

Worked example 3: Free fall

A stone is dropped from the top of a building and hits the ground after 3 s. Find the height of the building. (g = 10 m/s²)

List: u = 0 (dropped), a = g = 10 m/s², t = 3 s.

Use s = ut + ½at²: s = 0 + ½ × 10 × 3² = 45 m

Worked example 4: Thrown upwards

A ball is thrown vertically upwards with a velocity of 30 m/s. Find (a) the maximum height and (b) the time to reach it. (g = 10 m/s²)

Take upwards as positive, so a = −10 m/s² (gravity acts downwards). At maximum height, v = 0.

(a) v² = u² + 2as → 0 = 30² + 2(−10)s → 0 = 900 − 20s → s = 45 m

(b) v = u + at → 0 = 30 + (−10)t → t = 3 s

Choosing a direction as positive at the start, and sticking to it, prevents most sign errors.

Worked example 5: Unit conversion

An okada increases its speed uniformly from 18 km/h to 54 km/h in 5 s. Find its acceleration.

Convert to m/s first (divide km/h by 3.6):

• u = 18 ÷ 3.6 = 5 m/s
• v = 54 ÷ 3.6 = 15 m/s

a = (v − u) ÷ t = (15 − 5) ÷ 5 = 2 m/s²

Forgetting to convert km/h to m/s is one of the most common ways to lose marks in this topic.

Velocity–time graphs

Many questions present motion as a velocity–time graph. Two facts unlock them:

• The gradient (slope) gives the acceleration. A steeper line means greater acceleration, a horizontal line means constant velocity (zero acceleration), and a downward slope means deceleration.
• The area under the graph gives the distance (displacement).

Worked example 6

A car accelerates uniformly from rest to 12 m/s in 4 s, moves at 12 m/s for 6 s, then decelerates uniformly to rest in 2 s. Find the total distance travelled.

Split the area under the graph into shapes:

• Acceleration (triangle): ½ × 4 × 12 = 24 m
• Constant velocity (rectangle): 6 × 12 = 72 m
• Deceleration (triangle): ½ × 2 × 12 = 12 m

Total = 24 + 72 + 12 = 108 m

Common mistakes

• Not converting units: km/h to m/s, cm to m, minutes to seconds.
• Missing hidden information such as "from rest", "stops" or "maximum height".
• Sign errors with deceleration and vertical motion. Choose a positive direction and stick to it.
• Using the equations when acceleration isn't constant.
• Confusing distance and displacement in questions involving a change of direction.

Practise the method

The more problems you solve with the list–identify–choose method, the more automatic it becomes. Levia's Physics levels include motion questions from Foundations upwards, many with everyday Nigerian examples such as buses, okadas and canoes, and every answer shows the equation, the substitution and the units.