Using s = ut + 1/2 at squared in HSC Physics

A car is travelling east at \(12\text{ m s}^{-1}\). The driver brakes steadily for \(3.0\text{ s}\). You know the initial velocity, the acceleration, and the time, but you don’t know how far the car moves.

You might be tempted to calculate \(12 \times 3 = 36\text{ m}\). But that assumes the car keeps travelling at \(12\text{ m s}^{-1}\) for the whole three seconds. It doesn’t. It is slowing down.

The equation

\[
s = ut + \frac{1}{2}at^2
\]

fixes exactly this problem, provided the acceleration is constant.

The harder part in HSC questions is usually not substituting into the equation. It is deciding whether the equation applies at all, and choosing the signs of \(u\), \(a\), and \(s\) correctly.

01What problem does the equation solve?

Suppose an object starts with velocity \(u\) and then accelerates at a constant rate \(a\).

Its motion has two parts that we can picture.

First, imagine the object kept moving at its initial velocity for time \(t\). It would travel a displacement of

\[
ut.
\]

But acceleration changes its velocity while it moves. That change contributes some extra displacement, or removes some displacement if the acceleration points opposite the initial motion.

That acceleration contribution is

\[
\frac{1}{2}at^2.
\]

So the total displacement is

\[
s = ut + \frac{1}{2}at^2.
\]

Here:

• \(s\) is displacement, measured in metres (\(\text{m}\))
• \(u\) is initial velocity, measured in metres per second (\(\text{m s}^{-1}\))
• \(a\) is constant acceleration, measured in metres per second squared (\(\text{m s}^{-2}\))
• \(t\) is elapsed time, measured in seconds (\(\text{s}\))

Notice that \(s\) is displacement, not distance. That distinction becomes important whenever an object changes direction. If that distinction is still slippery, revise distance and displacement in HSC Physics before pushing further.

02Why is there a \(\frac{1}{2}\)?

The factor of \(\frac{1}{2}\) can look like something you just have to memorise. It isn’t arbitrary.

Imagine an object starts from rest and accelerates constantly at \(2\text{ m s}^{-2}\) for \(4\text{ s}\).

Its velocity increases like this:

• at \(0\text{ s}\): \(0\text{ m s}^{-1}\)
• at \(1\text{ s}\): \(2\text{ m s}^{-1}\)
• at \(2\text{ s}\): \(4\text{ m s}^{-1}\)
• at \(3\text{ s}\): \(6\text{ m s}^{-1}\)
• at \(4\text{ s}\): \(8\text{ m s}^{-1}\)

Predict the displacement. Would it be \(8 \times 4 = 32\text{ m}\)?

No. The object only reaches \(8\text{ m s}^{-1}\) at the end. It spends most of the four seconds travelling more slowly.

Because the velocity increases uniformly from \(0\) to \(8\text{ m s}^{-1}\), its average velocity is

\[
\frac{0 + 8}{2} = 4\text{ m s}^{-1}.
\]

So its displacement is

\[
s = v_{\text{av}}t = 4 \times 4 = 16\text{ m}.
\]

Using \(s = ut + \frac{1}{2}at^2\) gives the same result:

\[
\begin{aligned}
s &= ut + \frac{1}{2}at^2 \\
&= (0)(4) + \frac{1}{2}(2)(4^2) \\
&= 16\text{ m}.
\end{aligned}
\]

The \(\frac{1}{2}\) appears because, under constant acceleration, the velocity changes steadily rather than jumping instantly to its final value.

Another way to see this is with a velocity-time graph. The \(ut\) term is the rectangular area caused by the initial velocity. The \(\frac{1}{2}at^2\) term is the triangular area created by the changing velocity.

That picture is useful, but it has a limit: the neat triangle only appears when acceleration is constant. If acceleration changes with time, the velocity-time graph is generally not a straight line, and this equation cannot simply be used over the whole interval.

03The condition students forget: acceleration must be constant

The equation belongs to the constant-acceleration equations.

That means \(a\) must stay the same throughout the time interval you are analysing.

For example, these situations can reasonably be modelled with constant acceleration if the question tells you to do so:

• an object in free fall near Earth’s surface, with air resistance neglected
• a car accelerating uniformly along a straight road
• a train braking at a constant rate
• a trolley experiencing constant acceleration along a track

But suppose a car accelerates at \(2\text{ m s}^{-2}\) for two seconds and then at \(5\text{ m s}^{-2}\) for another two seconds.

Can you put \(t = 4\text{ s}\) into

\[
s = ut + \frac{1}{2}at^2
\]

using some convenient single value of \(a\)?

Not unless that value genuinely represents constant acceleration over the whole interval. In this example, it does not.

Instead, split the motion into intervals. Find the displacement during the first constant-acceleration interval, determine the new velocity, then analyse the second interval.

An acceleration-time graph can help you recognise when acceleration is constant and when the motion needs to be separated into sections.

04Choosing signs: choose a positive direction first

This is where many otherwise correct solutions fall apart.

Velocity and acceleration are vectors. In one-dimensional motion, their signs tell you which direction they point.

Before substituting anything, choose a positive direction.

For horizontal motion, you might choose:

\[
\text{east} = +
\]

which automatically means

\[
\text{west} = -.
\]

For vertical motion, you might choose:

\[
\text{up} = +
\]

which means

\[
\text{down} = -.
\]

You could choose the opposite convention and still get the right physics. The important thing is consistency.

Think of choosing the positive direction like deciding which end of a hallway gets the positive room numbers. Physics doesn’t care which end you choose. It cares that you don’t change your mind halfway down the corridor.

Signs do not mean “good”, “bad”, “speeding up”, or “slowing down”

A negative acceleration does not automatically mean an object is slowing down.

It means the acceleration points in the negative direction.

Suppose east is positive.

If a car is travelling east, then \(u > 0\).

If its acceleration points west, then \(a < 0\). The acceleration opposes the velocity, so the car slows down.

But suppose the car is travelling west. Then \(u < 0\). If its acceleration also points west, then \(a < 0\), and the car speeds up.

So the useful rule is:

The signs tell you direction first. Whether speed increases or decreases comes from comparing the directions of velocity and acceleration.

05A reliable sign-setting method

Before using the equation, write down your positive direction.

Then translate every vector quantity into that convention.

Suppose a ball is thrown upwards at \(15\text{ m s}^{-1}\), and you choose upwards as positive.

Then:

\[
u = +15\text{ m s}^{-1}
\]

and gravitational acceleration is downward, so

\[
a = -9.8\text{ m s}^{-2}.
\]

If the calculated displacement is \(s = +6.0\text{ m}\), the ball finishes \(6.0\text{ m}\) above its starting position.

If \(s = -6.0\text{ m}\), it finishes \(6.0\text{ m}\) below its starting position.

Do not attach a negative sign to acceleration merely because the object is “decelerating”. Attach the sign according to direction.

08Why \(t^2\) matters

Students sometimes remember the structure as “initial motion plus acceleration motion” but forget to square the time.

Check the units.

The first term has units

\[
ut = (\text{m s}^{-1})(\text{s}) = \text{m}.
\]

The second term has units

\[
at^2 = (\text{m s}^{-2})(\text{s}^2) = \text{m}.
\]

So both terms are displacements.

If you accidentally used \(\frac{1}{2}at\), its units would be

\[
(\text{m s}^{-2})(\text{s}) = \text{m s}^{-1},
\]

which is a velocity, not a displacement.

Unit checking will not catch every physics mistake, but it catches plenty of algebra mistakes cheaply.

09Why displacement grows with \(t^2\)

Suppose an object starts from rest with constant acceleration.

Then \(u = 0\), so

\[
s = \frac{1}{2}at^2.
\]

If you double the time, what happens to the displacement?

A tempting prediction is that the displacement doubles. But

\[
(2t)^2 = 4t^2,
\]

so the displacement becomes four times as large.

For example, an object starting from rest with acceleration \(3.0\text{ m s}^{-2}\) travels

\[
s = \frac{1}{2}(3.0)(2.0^2) = 6.0\text{ m}
\]

in \(2.0\text{ s}\).

After \(4.0\text{ s}\),

\[
s = \frac{1}{2}(3.0)(4.0^2) = 24\text{ m}.
\]

Twice the time gives four times the displacement.

Why? Because the object has longer to move and is moving faster during the later part of that longer interval.

10The most tempting misconception: negative displacement means moving backwards

Not necessarily.

Suppose east is positive. If an object has

\[
s = -5\text{ m},
\]

that tells you only that its final position is \(5\text{ m}\) west of where it started.

It could currently be moving west.

It could also have travelled west, turned around, and now be moving east while still remaining \(5\text{ m}\) west of its starting point.

Displacement describes a change in position. Velocity describes the rate and direction in which position is changing at that instant.

Those are connected ideas, but they are not interchangeable.

11A compact decision rule for HSC questions

Before using

\[
s = ut + \frac{1}{2}at^2,
\]

check four things.

1. Is the motion one-dimensional over the interval you are analysing?
   Choose one axis and define its positive direction.

2. Is acceleration constant?
   If not, this equation cannot be applied across the whole changing-acceleration interval.

3. Do you know \(u\), \(a\), and \(t\), or can you find them?
   This equation is particularly useful because final velocity does not appear.

4. Have you assigned signs by direction?
   Do this before substitution, not afterwards.

A useful habit is to write a line such as

\[
\text{upwards positive}
\]

before any vertical-motion calculation. It takes a few seconds and prevents a large fraction of sign errors.

12Questions and solutions

Question 1

A skateboarder travels north at \(3.0\text{ m s}^{-1}\) and accelerates uniformly north at \(0.80\text{ m s}^{-2}\) for \(5.0\text{ s}\). Calculate the skateboarder’s displacement during this interval.

Solution 1

The skateboarder moves \(\boxed{25\text{ m north}}\).

Take north as positive. Then

\[
u = +3.0\text{ m s}^{-1}, \qquad
a = +0.80\text{ m s}^{-2}, \qquad
t = 5.0\text{ s}.
\]

Using

\[
s = ut + \frac{1}{2}at^2,
\]

we get

\[
\begin{aligned}
s &= (3.0)(5.0) + \frac{1}{2}(0.80)(5.0^2) \\
&= 15 + 10 \\
&= 25\text{ m}.
\end{aligned}
\]

The positive result means the displacement is north. Both \(u\) and \(a\) are positive because they point in the chosen positive direction.

Question 2

A lift is moving upwards at \(2.4\text{ m s}^{-1}\). It accelerates uniformly downwards at \(0.60\text{ m s}^{-2}\) for \(3.0\text{ s}\).

Calculate its displacement during the \(3.0\text{ s}\) interval.

Solution 2

The lift’s displacement is \(\boxed{+4.5\text{ m}}\), so it finishes \(4.5\text{ m}\) above its initial position.

Choose upwards as positive. Then

\[
u = +2.4\text{ m s}^{-1}, \qquad
a = -0.60\text{ m s}^{-2}, \qquad
t = 3.0\text{ s}.
\]

Therefore,

\[
\begin{aligned}
s &= ut + \frac{1}{2}at^2 \\
&= (2.4)(3.0) + \frac{1}{2}(-0.60)(3.0^2) \\
&= 7.2 – 2.7 \\
&= 4.5\text{ m}.
\end{aligned}
\]

The acceleration is negative because it points downwards, not simply because the lift is slowing.

During these three seconds, the upward velocity and downward acceleration point in opposite directions, so the lift’s speed decreases.

Question 3

A small ball is projected vertically upwards from ground level at \(14.0\text{ m s}^{-1}\). Ignore air resistance and use \(g = 9.8\text{ m s}^{-2}\).

A student calculates its displacement after \(3.00\text{ s}\) as

\[
s = (14.0)(3.00) + \frac{1}{2}(9.8)(3.00^2)
\]

and obtains \(86.1\text{ m}\).

Identify the error, calculate the correct displacement, and state what the sign of your result means.

Solution 3

The student’s error is using the wrong sign for gravitational acceleration. The correct displacement is \(\boxed{-2.1\text{ m}}\), meaning the ball would be \(2.1\text{ m}\) below its release level after \(3.00\text{ s}\) if its motion were allowed to continue unobstructed.

If upwards is positive, then

\[
u = +14.0\text{ m s}^{-1}
\]

but gravity points downwards, so

\[
a = -9.8\text{ m s}^{-2}.
\]

Therefore,

\[
\begin{aligned}
s &= ut + \frac{1}{2}at^2 \\
&= (14.0)(3.00) + \frac{1}{2}(-9.8)(3.00^2) \\
&= 42.0 – 44.1 \\
&= -2.1\text{ m}.
\end{aligned}
\]

There is also an important physical assumption hidden in the calculation. The question says the ball was launched from ground level. A real ball would hit the ground before reaching a position \(2.1\text{ m}\) below it, unless there were a hole or drop beneath the launch point.

So the equation correctly describes the mathematical constant-acceleration trajectory, but the physical situation may end when the ball contacts the ground. Equations do not automatically know about barriers.

Question 4

A test vehicle begins at \(+6.0\text{ m s}^{-1}\), where east is positive.

For the first \(2.0\text{ s}\), it has constant acceleration \(-2.0\text{ m s}^{-2}\). For the next \(3.0\text{ s}\), its acceleration is constant at \(+1.0\text{ m s}^{-2}\).

A student proposes using

\[
s = ut + \frac{1}{2}at^2
\]

once over the full \(5.0\text{ s}\), with \(u = 6.0\text{ m s}^{-1}\) and an acceleration obtained by averaging \(-2.0\) and \(+1.0\).

Explain why this method is not valid, then calculate the vehicle’s total displacement.

Solution 4

The single-equation method is not valid because the acceleration is not constant over the full \(5.0\text{ s}\). The correct total displacement is \(\boxed{19.5\text{ m east}}\).

The motion must be split into the two intervals where acceleration is constant.

For the first interval,

\[
u_1 = +6.0\text{ m s}^{-1}, \qquad
a_1 = -2.0\text{ m s}^{-2}, \qquad
t_1 = 2.0\text{ s}.
\]

The displacement is

\[
\begin{aligned}
s_1 &= u_1t_1 + \frac{1}{2}a_1t_1^2 \\
&= (6.0)(2.0) + \frac{1}{2}(-2.0)(2.0^2) \\
&= 12 – 4 \\
&= 8.0\text{ m}.
\end{aligned}
\]

We also need the velocity at the end of this interval because it becomes the initial velocity for the second interval.

Using

\[
v = u + at,
\]

\[
\begin{aligned}
v_1 &= 6.0 + (-2.0)(2.0) \\
&= 2.0\text{ m s}^{-1}.
\end{aligned}
\]

For the second interval,

\[
u_2 = +2.0\text{ m s}^{-1}, \qquad
a_2 = +1.0\text{ m s}^{-2}, \qquad
t_2 = 3.0\text{ s}.
\]

So

\[
\begin{aligned}
s_2 &= u_2t_2 + \frac{1}{2}a_2t_2^2 \\
&= (2.0)(3.0) + \frac{1}{2}(1.0)(3.0^2) \\
&= 6.0 + 4.5 \\
&= 10.5\text{ m}.
\end{aligned}
\]

The total displacement is

\[
\begin{aligned}
s_{\text{total}} &= s_1 + s_2 \\
&= 8.0 + 10.5 \\
&= 18.5\text{ m}.
\end{aligned}
\]

Therefore,

\[
\boxed{s_{\text{total}} = 18.5\text{ m east}}.
\]

The important idea is that averaging the two acceleration values does not recreate the actual motion. The timing of each acceleration matters because the first interval changes the velocity entering the second interval.

13What this equation prepares you for

The equation

\[
s = ut + \frac{1}{2}at^2
\]

is more than a substitution formula. It links three ideas: an object’s initial motion, the way constant acceleration changes that motion, and the displacement accumulated over time.

Once the signs make sense, the other constant-acceleration equations become easier to organise. The same convention carries through: choose a positive direction, treat velocity and acceleration as signed quantities, and ask what the result physically means.

The next useful step is learning to move between these equations and motion graphs. A graph can show immediately when acceleration is constant, when an object changes direction, and when a single constant-acceleration equation stops being valid.