GCSE · Physics · Edexcel · Spec 1PH0

SUVAT v^2 - u^2 = 2ax equation

Starting from rest with a steady push, something reaches 10 m/s after 25 m. How far until it's doing 20 m/s? Your gut says 50 m. It's actually 100 m.

Speeding up from rest at a steady 2 m/s²: how fast after each metre?

025507510005.51116.522Distance travelled, x (m)Velocity, v (m/s)(50, 14.14)

Distance travelled, x (m): 50. Velocity, v (m/s): 14.14

Drag the dot along the curve. Try 25 m, then 50 m, then 100 m.

Watch out: Going twice as far does not make you twice as fast. At 25 m the object is doing 10 m/s. At 50 m it is only doing about 14.14 m/s. It has to go 100 m, four times as far, to reach 20 m/s.

The equation behind the curve

v² − u² = 2ax

Final velocity squared, minus initial velocity squared, equals 2 × acceleration × distance. v and u are in m/s, a is in m/s² and x is in m.

Two conditions. The acceleration must stay constant. And there's no time in it, so it's the equation to reach for when time is neither given nor asked for. Know any three of v, u, a and x and you can find the fourth.

Choosing the equation

Is this a job for v² − u² = 2ax?

Two questions decide it: is the acceleration constant, and is a time given or asked for? Sort each problem into the column that fits.

Still to sort

Use v² − u² = 2ax (0)

Constant acceleration, and no time given or asked for.

Where the line is: If a time turns up anywhere in the problem, as a given value or as the answer, the problem belongs in the next column instead.

Time is involved: use an equation with t in it (0)

A time is given, or a time is what you have to find.

Where the line is: This equation has no t in it. It can't use a time you're given, and it can't produce a time you're asked for.

Acceleration changes: this equation doesn't apply (0)

The acceleration isn't steady.

Where the line is: The equation only holds while the acceleration stays constant. If the acceleration changes, even careful substitution won't give you a trustworthy answer.

8 of 8 still to sort.

Rearranging: one balanced move at a time

v² − u² = 2ax

This is the starting point. Each rearrangement is one balanced move from here: whatever you do to one side, you do to the other.

1 / 5

Worked example: a car braking to a stop

Problem

A car travelling at 72 km/h brakes with a constant deceleration of 5 m/s². How far does it travel before it stops?

Spot the slip

Find the line where this answer goes wrong

A cyclist speeds up steadily from 2 m/s to 8 m/s over 15 m. Calculate her acceleration.

A student's answer — which line goes wrong?

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

The motion equation with no time in it, and the one where speed hides under a square.

What you need to know

  • v² − u² = 2ax links final velocity v, initial velocity u, acceleration a and distance travelled x.
  • It only works while the acceleration is constant.
  • It has no time in it: use it when time is neither given nor asked for.
  • Convert to SI units before substituting: m/s, m/s² and m.

The big picture

v² − u² = 2ax links final velocity, initial velocity, acceleration and distance for motion in a straight line with constant acceleration. It has no time in it, so it's the one to use when time is neither given nor asked for. Put everything in SI units, give a deceleration a negative sign, square each velocity on its own, and finish with a square root when you're finding v or u.

Key points

1From rest, u = 0 and v² = 2ax, so four times the distance gives only twice the speed.
2Coming to rest means v = 0.
3Slowing down while moving forwards makes a negative, and v ends up smaller than u.
4Rearranged forms: v² = u² + 2ax, u² = v² − 2ax, a = (v² − u²) ÷ 2x and x = (v² − u²) ÷ 2a.
5To find v or u, square-root the value you get for v² or u².

Worked example

Problem

A skateboarder rolling at 4 m/s speeds up with a constant acceleration of 1.5 m/s² over 11 m of slope. What is her velocity at the end?

⚠ Watch out

Working out (v − u)² instead of v² − u². With u = 4 m/s and v = 6 m/s, v² − u² = 36 − 16 = 20, but (v − u)² = 2² = 4. That's a completely different number. Square each velocity first, then subtract.

🧠

Memory hook

No t in sight? Reach for v² − u² = 2ax, and square before you subtract.

✓

Check yourself

A trolley slows from 6 m/s to a stop over 9 m. Before calculating anything: should a come out positive or negative? (Negative, because it's slowing down while moving forwards.)

Flashcards

(13)
What four quantities does v² − u² = 2ax link?
Final velocity v, initial velocity u, acceleration a and distance travelled x.
What must be true about the acceleration for v² − u² = 2ax to work?
It must be constant (uniform). If the acceleration changes, the equation doesn't apply.
Which quantity does v² − u² = 2ax leave out, and when does that make it the right choice?
Time. Use it when time is neither given nor asked for.
What SI units do v, u, a and x need before you substitute?
v and u in m/s, a in m/s², x in m.
Rearrange v² − u² = 2ax to make v² the subject.
v² = u² + 2ax
Rearrange v² − u² = 2ax to make u² the subject.
u² = v² − 2ax
Rearrange v² − u² = 2ax to make a the subject.
a = (v² − u²) ÷ 2x
Rearrange v² − u² = 2ax to make x the subject.
x = (v² − u²) ÷ 2a
You've worked out v². What's the last step to get v?
Take the square root.
An object starts from rest. What does that tell you, and what does the equation become?
u = 0, so v² = 2ax.
An object comes to rest. Which value do you know?
v = 0
An object slows down while moving forwards. What sign does a take?
Negative. v ends up smaller than u.
Starting from rest at a constant acceleration, how much further must you go to double your speed?
Four times as far, because v² is proportional to x.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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