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?
Drag the dot along the curve. Try 25 m, then 50 m, then 100 m.
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.
Rearranging: one balanced move at a time
This is the starting point. Each rearrangement is one balanced move from here: whatever you do to one side, you do to the other.
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?
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
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?
What must be true about the acceleration for v² − u² = 2ax to work?
Which quantity does v² − u² = 2ax leave out, and when does that make it the right choice?
What SI units do v, u, a and x need before you substitute?
Rearrange v² − u² = 2ax to make v² the subject.
Rearrange v² − u² = 2ax to make u² the subject.
Rearrange v² − u² = 2ax to make a the subject.
Rearrange v² − u² = 2ax to make x the subject.
You've worked out v². What's the last step to get v?
An object starts from rest. What does that tell you, and what does the equation become?
An object comes to rest. Which value do you know?
An object slows down while moving forwards. What sign does a take?
Starting from rest at a constant acceleration, how much further must you go to double your speed?
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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