What Are the Equations of Motion?

🧠 What Are the Equations of Motion? A Friendly A Level Guide

⚙️ What we’re doing (and why it matters)

Right, quick truth: SUVAT isn’t a topic on its own. It shows up everywhere — projectiles, constant acceleration motion, even mixed in with Newton’s Laws questions. That’s why AQA, Edexcel and OCR keep returning to it.

We’re talking motion in a straight line with constant acceleration. Not “sort of constant.” Constant. If acceleration changes, SUVAT stops helping and we switch to a different toolkit.

Before any calculation, draw a tiny diagram. Always. Sketching where t

🔙 Previous topic:

“Revisit why mechanics challenges you before moving into engineering.”

🧠 The five letters you must own

You’ll see these symbols in every mechanics course:

u = initial velocity
v = final velocity
a = acceleration (constant)
t = time
s = displacement (signed distance from the start)

I tell my classes: write them in that order at the top of the page, then fill what the question gives you. The empty boxes tell you which equation to use.

📏 The five SUVAT equations

Keep it readable, like you’d say it out loud.

$v = u + a t$
$v^2 = u^2 + 2 a s$
$s = \frac{(u + v)t}{2}$
$s = u t + \frac{1}{2} a t^2$
$s = v t – \frac{1}{2} a t^2$

You never use all five at once. Pick the one that doesn’t include the letter you don’t know and don’t need.

✅ Exam tip: If time $t$ is missing from the data, try $v^2 = u^2 + 2 a s$ first.

❗ Common slip: Using a formula that contains a letter you don’t know and don’t have a second equation for. Choose smarter, not harder.

✏️ Example 1 — Straightforward constant acceleration

A particle accelerates at 1.5 m/s^2 for 12 s and finishes at 25 m/s. Find the initial velocity and the distance covered.

Known: a = 1.5, t = 12, v = 25. Unknowns: u and s.

Find u first using v = u + a t
25 = u + 1.5 × 12 → u = 25 − 18 = 7 m/s.

Then s using s = u t + (1/2) a t^2
s = 7 × 12 + 0.5 × 1.5 × 12^2
s = 84 + 0.75 × 144 = 84 + 108 = 192 m.

🧠 Teacher aside: I always solve for u before s where I can. It tidy-ups the numbers and stops random substitution mistakes.

📈 Vertical motion under gravity (the sign game)

For vertical motion we usually take upwards as positive. Gravity is then a = −9.8 m/s^2 (some boards use 9.81; either is fine if you’re consistent).

If something is dropped, u = 0.
If it’s thrown up, u > 0 and the acceleration is still −9.8.
If it’s thrown down, u is negative or you flip your positive direction. Just be consistent.

✅ Exam tip: Write “up is +” or “down is +” on your diagram. Saves sign errors later.
❗ Common slip: Mixing sign conventions halfway through the working. Pick one, stick to it.

✏️ Example 2 — Coin dropped from a height

A coin is dropped from rest at the top of a 12 m building. Take a = 9.8 m/s^2 downward. Find the time to hit the ground and the speed of impact.

Let’s call downward positive to keep numbers friendly.
u = 0, a = +9.8, s = +12.

Use s = u t + (1/2) a t^2
12 = 0 × t + 0.5 × 9.8 × t^2 → t^2 = 24 / 9.8 → t ≈ 1.56 s.

Speed at impact from v^2 = u^2 + 2 a s
v^2 = 0 + 2 × 9.8 × 12 = 235.2 → v ≈ 15.3 m/s downward.

🧠 Teacher aside: If you had taken up as positive, you’d get the same speed with a negative sign telling you the direction — perfectly fine.

📊 Two-interval timing (classic Edexcel style)

You time a car between lamp posts. The posts are 30 m apart. As the car speeds up, the times for two consecutive gaps are 1.2 s then 1.0 s. Assume constant acceleration.
a) Find the acceleration and speed at the first post.
b) How long did the car take to travel the 30 m before the first post?

Let the first post be A, then B, then C.

From A to B:
s = 30, time = 1.2, unknowns u and a.
s = u t + (1/2) a t^2 → 30 = 1.2 u + 0.72 a … (1)

From A to C together (avoid extra unknown at B):
s = 60, time = 2.2.
60 = 2.2 u + (1/2) a (2.2)^2 → 60 = 2.2 u + 2.42 a … (2)

Solve (1) and (2): a ≈ 4.55 m/s^2, u ≈ 22.3 m/s at A.

Part b (the 30 m before A):
Now we know the speed at A (22.3 m/s) and the same acceleration (4.55 m/s^2). We want the time t for 30 m before A, so the car is slightly slower than 22.3 in that interval.

Use s = v t − (1/2) a t^2 with v = 22.3 (speed at the end of the interval), a = 4.55, s = 30:
30 = 22.3 t − 2.275 t^2 → 2.275 t^2 − 22.3 t + 30 = 0.

Solve the quadratic: t ≈ 1.61 s (the larger root doesn’t fit the context).

✅ Exam tip: When two consecutive timings are given, avoid inventing an extra unknown velocity at the middle post. Jump A to C to keep it to two equations.
❗ Common slip: Using s = (u + v) t / 2 when you don’t yet know v. You’ll box yourself in.

🌧️ Free fall basics (and typical traps)

When objects move under gravity alone:

Use a = 9.8 m/s^2 and choose a positive direction.
On the way up, acceleration is negative if up is positive.
At the top of the arc, v = 0. That’s a great equation starter.

Students often forget that on the way up, velocity is still positive (until the top), but acceleration is negative all the time. That mismatch is normal.

✏️ Example 3 — Thrown vertically down

A brick is thrown straight down from a building with an initial speed 1.5 m/s. Height is 19.5 m. Find:
a) impact speed
b) time of fall

Take down as positive.
u = +1.5, a = +9.8, s = +19.5.

a) v^2 = u^2 + 2 a s → v^2 = 1.5^2 + 2 × 9.8 × 19.5 = 380.25 → v ≈ 19.5 m/s downward.
b) Use v = u + a t → 19.5 = 1.5 + 9.8 t → t ≈ 1.84 s.

✅ Exam tip: If you’re short on time, go straight to v^2 = u^2 + 2 a s to get the speed. No need to find t first.
❗ Common slip: Treating “downward throw” as zero initial speed. It isn’t — that’s a different motion.

✏️ Example 4 — Thrown up, returns through a height twice

A particle is projected vertically upwards at 34.3 m/s. When is it 49 m above the start point? (first time and second time)

Up is positive.
u = 34.3, a = −9.8, s = +49.

s = u t + (1/2) a t^2 → 49 = 34.3 t − 4.9 t^2
Rearrange: 4.9 t^2 − 34.3 t + 49 = 0.

Solve the quadratic: t = 2.0 s and t = 5.0 s (to 1 dp, they’re exact here).
First time on the way up, second time on the way down.

🧠 Teacher aside: Two times for the same height is a classic shape question. Same level up and down — same displacement, different velocities.

🔍 How to pick the right SUVAT every time

Here’s the quick routine I drill into classes:

Write u, v, a, t, s in a row and fill what you know.
Circle the unknown you want.
Pick the formula that doesn’t include the letter you don’t have.
If two unknowns remain, look for a second interval or a second equation.
Keep the sign convention visible on your sketch.

This sounds simple, but under pressure students skip step 1 and try to “remember the right one.” Don’t. The table chooses it for you.

✅ Exam habits that score (AQA / Edexcel / OCR)

AQA: loves wording like “smooth” (no friction) and “particle” (size ignored). Watch the sign of acceleration in vertical motion.
Edexcel: enjoys back-to-back intervals (lamp posts, beacons, markers). Jump the middle if it introduces an extra unknown.
OCR: will often test interpretation — is the object still rising? Is velocity positive or negative there? Say it in words as well as numbers.

Always include units. A correct number with no m, s or m/s^2 can still drop a mark.

❗ Common mistakes (and how to fix them)

Calling distance “displacement” (they’re not the same).
Swapping signs mid-solution — choose up or down as positive and stick to it.
Using a formula with a variable you don’t know and don’t have a way to find.
Forgetting that v = 0 at the top of a vertical throw — it unlocks the time or height in one line.
Ignoring the second time a level is reached on the way down.

If you spot one of these in your work, correct it and write the fix in words. Examiners love seeing and thinking.

🧾 Quick recap (the essentials)

Idea	What to remember
Constant acceleration	SUVAT applies; if a varies, stop using it
u, v, a, t, s	Fill a table first — it chooses your formula
Graph sense	Two times at same height: up and down
Gravity	Use 9.8 m/s^2; watch the sign
Equations	Pick the one that drops the letter you don’t need

🧠 Teacher reflection: I’ve seen students double their marks just by drawing a sketch and writing u, v, a, t, s every single time. It slows you down for 20 seconds and saves you minutes of confusion.

🚀 Next Steps

If you want to turn this understanding into exam confidence, head to our 👉 live online A Level Maths sessions step by step, with teacher walkthroughs, practice sets, and clear solutions you can trust.

From SUVAT to Newton’s Laws to projectiles, it’s the same logic reused. Once that penny drops, the marks follow.

Author Bio

S. Mahandru • Head of Maths, Exam.tips

S. Mahandru is Head of Maths at Exam.tips. With over 15 years of experience, he simplifies complex calculus topics and provides clear worked examples, strategies, and exam-focused guidance.

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What Are the Equations of Motion?