Using v² = u² + 2as in HSC Physics: Assumptions and Examples

A car is braking towards a red light. You know its velocity when the brakes are applied, you know its acceleration, and you know how far away the line is. There is one awkward problem: nobody measured the braking time.

Can you still decide whether the car stops before the line?

You can, provided the acceleration is constant. The equation

\[
v^2 = u^2 + 2as
\]

connects velocity, acceleration, and displacement without needing time. That makes it especially useful when the question gives distances but no time information.

Before using it, though, there are two traps to deal with. The \(s\) is displacement, not distance, and the equation assumes constant acceleration. Ignore either of those details and a perfectly tidy calculation can give a physically wrong answer.

01What the equation is really saying

Imagine an object travelling along a straight road. Every second, its velocity changes by the same amount.

If it starts at \(12\text{ m s}^{-1}\) and has an acceleration of \(-3\text{ m s}^{-2}\), its velocity goes:

• \(12\text{ m s}^{-1}\)
• \(9\text{ m s}^{-1}\)
• \(6\text{ m s}^{-1}\)
• \(3\text{ m s}^{-1}\)
• \(0\text{ m s}^{-1}\)

The minus sign does not mean the object is necessarily moving backwards. It tells us that the acceleration points in the negative direction we chose.

While the velocity is changing, the object is also moving through some displacement. The equation \(v^2 = u^2 + 2as\) links those two changes.

The symbols mean:

“Initial” and “final” just mean the beginning and end of the interval you are studying. They do not have to mean the beginning and end of the object’s entire journey.

02Why there is no time in the equation

Calling this the “time-free” equation can be slightly misleading. Time has not stopped existing. We have simply removed it algebraically.

For constant acceleration,

\[
v = u + at
\]

where \(t\) is the elapsed time in seconds. If you need to refresh that relationship, see The Equation \(v = u + at\): Constant Acceleration in HSC Physics.

Constant acceleration also means the velocity changes uniformly, so the average velocity during the interval is

\[
v_{\text{avg}} = \frac{u+v}{2}.
\]

Therefore the displacement is

\[
s = \frac{u+v}{2}t.
\]

Now look at the first equation again:

\[
v-u=at.
\]

Multiply \(v-u\) by \(v+u\):

\[
(v-u)(v+u)=at(v+u).
\]

The left side is a difference of two squares:

\[
v^2-u^2=at(v+u).
\]

From the displacement equation,

\[
t(u+v)=2s.
\]

So

\[
v^2-u^2=2as,
\]

which gives

\[
\boxed{v^2=u^2+2as}.
\]

The important point is not the algebra itself. Both equations used in the derivation depend on constant acceleration. That assumption is built into the result.

03When should you use \(v^2=u^2+2as\)?

A useful decision rule is simple: look at the variables in the question.

The equation contains \(u\), \(v\), \(a\), and \(s\), but not \(t\).

So it is often the best choice when:

• acceleration is constant,
• the motion is along one dimension, or you are analysing one component of motion,
• you know three of \(u\), \(v\), \(a\), and \(s\),
• and time is unknown or unnecessary.

It is usually a poor choice if time is the quantity you need directly.

For example, suppose you know \(u\), \(a\), and \(t\), and need \(v\). Using \(v=u+at\) is much cleaner. There is no prize for using the most complicated equation available.

04Signs matter more than memorising the formula

Suppose a car travels east and slows down. If we choose east as positive, then:

• its velocity is positive,
• its displacement while still travelling east is positive,
• its acceleration is negative.

That gives a term \(2as\) that is negative, so \(v^2\) becomes smaller than \(u^2\). That makes physical sense because the car is slowing down.

Now predict what happens if the acceleration and displacement have the same sign.

Then \(as\) is positive, so

\[
v^2>u^2.
\]

The magnitude of the velocity increases.

The equation handles speeding up and slowing down automatically, but only if the signs describe the actual directions correctly.

If displacement and distance are easy to mix up, review Distance and Displacement in HSC Physics Explained Clearly. In this equation, \(s\) is always the signed displacement along your chosen axis.

05The square on \(v\) creates an important trap

Suppose you rearrange the equation to find the final velocity:

\[
v^2=u^2+2as.
\]

You might then write

\[
v=\sqrt{u^2+2as}.
\]

That is incomplete.

Mathematically,

\[
v=\pm\sqrt{u^2+2as}.
\]

The equation gives you the magnitude of the velocity, but the physical situation may be needed to determine its direction.

This matters especially in vertical motion.

06The biggest assumption: acceleration must be constant

A tempting idea is that you can use an average acceleration whenever the actual acceleration changes.

Usually, you cannot.

Consider two objects that both start from rest and reach \(4\text{ m s}^{-1}\) after \(2\text{ s}\).

Object A accelerates constantly at \(2\text{ m s}^{-2}\). Its displacement is

\[
s=ut+\frac12at^2
=0+\frac12(2)(2^2)
=4\text{ m}.
\]

Object B instead accelerates at \(4\text{ m s}^{-2}\) for the first second, then travels at its new constant velocity of \(4\text{ m s}^{-1}\) for another second.

During its first second,

\[
s_1=\frac12(4)(1^2)=2\text{ m}.
\]

During its second second,

\[
s_2=(4)(1)=4\text{ m}.
\]

Its total displacement is therefore

\[
s=2+4=6\text{ m}.
\]

Both objects have the same initial velocity, final velocity, elapsed time, and average acceleration of \(2\text{ m s}^{-2}\). Yet one travels \(4\text{ m}\) and the other travels \(6\text{ m}\).

Why?

Because displacement depends on the entire velocity history. Object B reaches its high velocity earlier, so it spends more time travelling quickly.

Using an average acceleration in

\[
v^2=u^2+2as
\]

would incorrectly predict the same displacement for both objects.

07A quick test before using the equation

Before substituting numbers, check these points:

That last point is about efficiency rather than validity. The equation may still be true when time is known. It is simply most useful when time is absent.

08What if \(a=0\)?

There is a useful edge case.

If acceleration is zero, then \(v=u\). The equation becomes

\[
v^2=u^2.
\]

That is true, but it tells you nothing about the displacement.

For example, an object travelling at a constant \(7\text{ m s}^{-1}\) might travel \(7\text{ m}\), \(70\text{ m}\), or \(700\text{ m}\), depending on how long it moves.

So an equation can be valid without being useful for finding a particular unknown.

When \(a=0\), displacement is normally found from

\[
s=ut,
\]

provided the elapsed time \(t\) is known.

09Questions and solutions

Question 1

A skateboarder is moving at \(5.0\text{ m s}^{-1}\) along a straight path. They accelerate constantly at \(1.5\text{ m s}^{-2}\) through a displacement of \(8.0\text{ m}\). Find their final velocity.

Solution 1

The final velocity is \(7.0\text{ m s}^{-1}\) in the original direction of motion.

Take the original direction as positive. Then

\[
u=5.0\text{ m s}^{-1},\qquad
a=1.5\text{ m s}^{-2},\qquad
s=8.0\text{ m}.
\]

Using

\[
v^2=u^2+2as,
\]

we get

\[
\begin{aligned}
v^2
&=(5.0)^2+2(1.5)(8.0)\\
&=25+24\\
&=49\text{ m}^2\text{ s}^{-2}.
\end{aligned}
\]

Therefore,

\[
v=7.0\text{ m s}^{-1}.
\]

Mathematically the square root also allows \(-7.0\text{ m s}^{-1}\), but that direction does not fit this motion. The skateboarder begins moving forwards and has a forwards acceleration, so there is nothing here that could reverse their direction.

Question 2

A small electric cart is travelling at \(10.0\text{ m s}^{-1}\). Over a displacement of \(21.0\text{ m}\) in the same direction, its velocity decreases to \(4.0\text{ m s}^{-1}\). Assuming constant acceleration, calculate the acceleration.

Solution 2

The acceleration is \(-2.00\text{ m s}^{-2}\).

Take the cart’s direction of travel as positive. Then

\[
u=10.0\text{ m s}^{-1},\qquad
v=4.0\text{ m s}^{-1},\qquad
s=21.0\text{ m}.
\]

Rearrange

\[
v^2=u^2+2as
\]

to give

\[
a=\frac{v^2-u^2}{2s}.
\]

Substituting,

\[
\begin{aligned}
a
&=\frac{(4.0)^2-(10.0)^2}{2(21.0)}\\
&=\frac{16-100}{42}\\
&=-2.00\text{ m s}^{-2}.
\end{aligned}
\]

The negative sign means the acceleration points opposite the chosen positive direction. That matches the observation that the cart is slowing while still moving forwards.

Question 3

A particle starts at \(x=0\) with velocity \(+6.0\text{ m s}^{-1}\) and constant acceleration \(-2.0\text{ m s}^{-2}\). What velocity or velocities can it have when it is at \(x=+5.0\text{ m}\)?

Explain why there may be more than one answer.

Solution 3

The particle can have either \(+4.0\text{ m s}^{-1}\) or \(-4.0\text{ m s}^{-1}\) at \(x=+5.0\text{ m}\).

Here,

\[
u=+6.0\text{ m s}^{-1},\qquad
a=-2.0\text{ m s}^{-2},\qquad
s=+5.0\text{ m}.
\]

Using

\[
v^2=u^2+2as,
\]

gives

\[
\begin{aligned}
v^2
&=(6.0)^2+2(-2.0)(5.0)\\
&=36-20\\
&=16\text{ m}^2\text{ s}^{-2}.
\end{aligned}
\]

So

\[
v=\pm4.0\text{ m s}^{-1}.
\]

Both signs are physically possible.

The particle initially travels in the positive direction while the negative acceleration slows it. Its stopping displacement is

\[
\begin{aligned}
0^2&=(6.0)^2+2(-2.0)s\\
s&=9.0\text{ m}.
\end{aligned}
\]

It therefore passes \(x=5.0\text{ m}\) once on its way towards \(x=9.0\text{ m}\), with velocity \(+4.0\text{ m s}^{-1}\). After stopping and reversing, it passes \(x=5.0\text{ m}\) again with velocity \(-4.0\text{ m s}^{-1}\).

The tempting mistake is to take only the positive square root. The squared-velocity equation does not retain enough information to distinguish the two passes through the same position.

Question 4

An object starts at \(2.0\text{ m s}^{-1}\). For the first \(1.0\text{ s}\), it accelerates at \(6.0\text{ m s}^{-2}\). It then continues for another \(2.0\text{ s}\) with zero acceleration.

A student calculates the average acceleration over the whole \(3.0\text{ s}\) interval and substitutes it into \(v^2=u^2+2as\) to find the total displacement.

Explain why this method is invalid, and calculate the actual total displacement.

Solution 4

The method is invalid because the acceleration is not constant, and the actual total displacement is \(21\text{ m}\).

During the first second,

\[
u=2.0\text{ m s}^{-1},\qquad
a=6.0\text{ m s}^{-2},\qquad
t=1.0\text{ s}.
\]

The velocity after that second is

\[
v=u+at
=2.0+(6.0)(1.0)
=8.0\text{ m s}^{-1}.
\]

The first displacement is

\[
\begin{aligned}
s_1
&=ut+\frac12at^2\\
&=(2.0)(1.0)+\frac12(6.0)(1.0)^2\\
&=5.0\text{ m}.
\end{aligned}
\]

For the next \(2.0\text{ s}\), the object travels at a constant \(8.0\text{ m s}^{-1}\), so

\[
s_2=vt=(8.0)(2.0)=16.0\text{ m}.
\]

Therefore,

\[
s_{\text{total}}=5.0+16.0=21.0\text{ m}.
\]

The average acceleration over the three seconds is

\[
a_{\text{avg}}=\frac{8.0-2.0}{3.0}
=2.0\text{ m s}^{-2}.
\]

If that value were incorrectly substituted into the constant-acceleration equation,

\[
\begin{aligned}
8.0^2&=2.0^2+2(2.0)s\\
64&=4+4s\\
s&=15\text{ m}.
\end{aligned}
\]

That is \(6\text{ m}\) too small.

The trap is thinking that the average acceleration preserves everything important about the journey. It does not. The real object reaches \(8.0\text{ m s}^{-1}\) after only one second and then spends two whole seconds travelling at that higher velocity.

Question 5

A research trolley travels along a straight track at a constant velocity of \(7.0\text{ m s}^{-1}\). A student substitutes \(a=0\) and \(v=u=7.0\text{ m s}^{-1}\) into

\[
v^2=u^2+2as
\]

and obtains

\[
49=49.
\]

They conclude that the displacement must be zero.

Explain the mistake. If the trolley travels for \(12.0\text{ s}\), find its displacement.

Solution 5

The displacement is \(84\text{ m}\), and the equation \(v^2=u^2+2as\) does not determine \(s\) when \(a=0\).

With zero acceleration,

\[
v=u.
\]

Substituting \(a=0\) gives

\[
v^2=u^2+0,
\]

so the equation reduces to

\[
v^2=u^2.
\]

The displacement has disappeared from the equation. The statement \(49=49\) is an identity, not evidence that \(s=0\).

Because the velocity is constant, use

\[
s=vt.
\]

Substituting

\[
v=7.0\text{ m s}^{-1},\qquad
t=12.0\text{ s},
\]

gives

\[
s=(7.0)(12.0)=84\text{ m}.
\]

This is an important general lesson: an equation can be completely valid but still contain no useful information about the quantity you are trying to find.

10What this equation prepares you to do next

The real skill is not remembering \(v^2=u^2+2as\). It is recognising the structure of the motion.

If acceleration is constant and time is missing, the equation can connect a change in velocity directly to displacement. It also forces you to think carefully about direction, signs, and what a squared velocity can and cannot tell you.

The next useful step is to connect these equations to motion graphs. Constant acceleration appears as a constant value on an acceleration-time graph, a straight line on a velocity-time graph, and curved motion on a displacement-time graph. Once those representations agree with the equations, kinematics becomes much less about choosing formulas and much more about describing one physical story in several different ways.