The Equation v = u + at: Constant Acceleration in HSC Physics

A car is moving at \(12\ \text{m s}^{-1}\). Three seconds later, it is moving at \(21\ \text{m s}^{-1}\). You could treat those as two unrelated velocity measurements. But if the acceleration stayed constant between them, you can predict every velocity in between with one short equation:

\[
v = u + at
\]

Before worrying about the letters, make a prediction. If the car gains \(3\ \text{m s}^{-1}\) of velocity every second, what should happen after 3 seconds?

It should gain \(9\ \text{m s}^{-1}\). Starting at \(12\ \text{m s}^{-1}\), that gives \(21\ \text{m s}^{-1}\).

That is all \(v = u + at\) is doing. The equation keeps track of a starting velocity, then adds the velocity change produced by a constant acceleration.

01Start with velocity change

Acceleration tells you how quickly velocity changes.

If an object has an acceleration of \(2\ \text{m s}^{-2}\), its velocity changes by \(2\ \text{m s}^{-1}\) every second.

So if it starts at \(5\ \text{m s}^{-1}\):

The pattern is:

\[
\text{new velocity} = \text{starting velocity} + \text{velocity change}
\]

For constant acceleration, the velocity change is:

\[
\Delta v = at
\]

where:

• \(\Delta v\) is the change in velocity, measured in \(\text{m s}^{-1}\)
• \(a\) is the constant acceleration, measured in \(\text{m s}^{-2}\)
• \(t\) is the elapsed time, measured in seconds

If the starting velocity is \(u\) and the final velocity is \(v\), then:

\[
v = u + at
\]

Here:

• \(u\) is the initial velocity, in \(\text{m s}^{-1}\)
• \(v\) is the final velocity, in \(\text{m s}^{-1}\)
• \(a\) is the constant acceleration, in \(\text{m s}^{-2}\)
• \(t\) is the elapsed time, in seconds

The letters matter less than the structure:

\[
\boxed{\text{final velocity} = \text{initial velocity} + \text{change in velocity}}
\]

02Why does \(at\) give a velocity change?

The units tell you.

Acceleration has units of \(\text{m s}^{-2}\). Multiplying by time gives:

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

So \(at\) has the units of velocity.

For example, an acceleration of \(4\ \text{m s}^{-2}\) acting for \(2.5\ \text{s}\) produces:

\[
\Delta v = at = (4)(2.5) = 10\ \text{m s}^{-1}
\]

That does not mean the final velocity is automatically \(10\ \text{m s}^{-1}\). It means the velocity has changed by \(10\ \text{m s}^{-1}\).

If the object started at \(6\ \text{m s}^{-1}\), it finishes at \(16\ \text{m s}^{-1}\).

If it started at \(-8\ \text{m s}^{-1}\), it finishes at \(2\ \text{m s}^{-1}\).

That distinction between velocity and velocity change is one of the main places students lose track of what the equation is saying.

03Where the equation comes from

The average acceleration over a time interval is defined as:

\[
a = \frac{v-u}{t}
\]

If the acceleration is constant, that same value of \(a\) applies throughout the interval.

Multiply both sides by \(t\):

\[
at = v-u
\]

Then add \(u\) to both sides:

\[
v = u+at
\]

So \(v=u+at\) is not a separate rule that needs to be memorised without meaning. It is the constant-acceleration definition rearranged.

This also tells you its limitation: you may only use it directly when acceleration is constant over the time interval you are considering.

If the acceleration changes with time, the simple product \(at\) no longer necessarily gives the velocity change. In that situation, you need to consider the area under an acceleration-time graph. That is developed further in tutorgum’s guide to acceleration-time graphs.

04The most important setup decision: choose a positive direction

Suppose a cyclist is moving east and slowing down.

A common mistake is to say:

“It’s slowing down, so its velocity must be negative.”

Not necessarily.

The sign of velocity tells you direction. It does not tell you whether the object is speeding up or slowing down.

You choose which direction is positive before doing the calculation.

For example, choose east as positive. Then a cyclist moving east has positive velocity. If the cyclist is slowing while still moving east, the acceleration points west, so the acceleration is negative.

You might have:

\[
u = +14\ \text{m s}^{-1}, \qquad a=-2.0\ \text{m s}^{-2}
\]

After \(3.0\ \text{s}\):

\[
v = 14 + (-2.0)(3.0) = 8.0\ \text{m s}^{-1}
\]

The positive final velocity tells you the cyclist is still moving east. The smaller magnitude tells you the cyclist has slowed down.

If you are not comfortable with signed velocities yet, the distinction between speed and velocity is worth fixing first. See Speed and Velocity: HSC Physics Explained Clearly.

05A reliable setup routine

When you see a constant-acceleration question involving initial velocity, final velocity, acceleration, and time, use this sequence.

1. Choose a positive direction.
2. Write the known quantities with signs and units.
3. Identify the unknown.
4. Use \(v=u+at\), or rearrange it if necessary.
5. Substitute values only after the equation is arranged.
6. Interpret the sign of the answer physically.

The last step is not decoration. In velocity questions, the sign can tell you whether the object has reversed direction.

06Negative acceleration does not always mean slowing down

Imagine two people walking towards each other in a corridor. You decide that right is positive.

One person is moving to the right, so their velocity is positive. The other is moving to the left, so their velocity is negative. Nothing about those signs tells you who is speeding up.

Acceleration works the same way.

An object speeds up when its acceleration points in the same direction as its velocity.

It slows down when acceleration points in the opposite direction to its velocity.

So all four combinations are possible:

That third row often feels strange. If both velocity and acceleration are negative, the object is moving in the negative direction and accelerating further in that same direction. Its velocity might change from \(-3\ \text{m s}^{-1}\) to \(-7\ \text{m s}^{-1}\). The number became “more negative”, but the speed increased from \(3\ \text{m s}^{-1}\) to \(7\ \text{m s}^{-1}\).

The minus sign is about direction, not bad vibes.

07Rearranging \(v=u+at\)

You will not always be asked for \(v\).

The same equation can find any one of its four quantities if the other three are known.

Starting with:

\[
v=u+at
\]

you can obtain:

\[
u=v-at
\]

\[
a=\frac{v-u}{t}
\]

and:

\[
t=\frac{v-u}{a}
\]

Do not try to memorise four separate equations. Start from \(v=u+at\) and rearrange it.

There is also a useful physical check hidden in the last form:

\[
t=\frac{\text{velocity change}}{\text{velocity change per second}}
\]

That is exactly what you would expect time to mean.

08The trap hidden in “comes to rest”

Suppose a question says an object “comes to rest”.

That gives you:

\[
v=0
\]

It does not automatically mean:

\[
a=0
\]

An object can have zero velocity at one instant while still accelerating.

A ball thrown vertically upwards is the standard example. At the top of its flight, its vertical velocity is momentarily zero, but its acceleration due to gravity is still directed downwards.

Similarly, the cart in the earlier example had \(v=0\) after \(2.5\ \text{s}\), but its acceleration was still \(-2.0\ \text{m s}^{-2}\).

Velocity describes motion at an instant. Acceleration describes how velocity is changing at that instant. They are different quantities.

09When you should not use \(v=u+at\)

The equation assumes constant acceleration.

Consider a car whose acceleration is:

• \(1\ \text{m s}^{-2}\) during the first second,
• \(4\ \text{m s}^{-2}\) during the second second,
• \(0\ \text{m s}^{-2}\) during the third second.

There is no single constant value of \(a\) that describes the whole three-second interval.

It would therefore be wrong to choose one of those accelerations and write:

\[
v=u+at
\]

for the full interval.

Could you use the equation separately during each interval where acceleration is constant? Yes. You could calculate the velocity change one section at a time.

Alternatively, the total velocity change is the area under the acceleration-time graph.

This shows something deeper about \(v=u+at\). On an acceleration-time graph with constant acceleration, the graph is a horizontal line. The area underneath is a rectangle:

\[
\text{area}=a\times t
\]

and that area is:

\[
\Delta v=at
\]

So the equation is really a special case of a broader idea about accumulation.

10Questions and solutions

Question 1

A skateboarder is travelling east at \(4.0\ \text{m s}^{-1}\). She accelerates uniformly east at \(1.5\ \text{m s}^{-2}\) for \(6.0\ \text{s}\). Find her final velocity.

Solution 1

The skateboarder’s final velocity is \(\boxed{13\ \text{m s}^{-1}\text{ east}}\).

Choose east as positive:

\[
u=+4.0\ \text{m s}^{-1}, \qquad a=+1.5\ \text{m s}^{-2}, \qquad t=6.0\ \text{s}
\]

Using:

\[
v=u+at
\]

gives:

\[
v=4.0+(1.5)(6.0)
=4.0+9.0
=13\ \text{m s}^{-1}
\]

The positive sign means the skateboarder is moving east. Her velocity increased by \(9.0\ \text{m s}^{-1}\) during the six seconds.

Question 2

A robot moves along a straight track at \(11\ \text{m s}^{-1}\) to the right. It experiences a constant acceleration of \(2.5\ \text{m s}^{-2}\) to the left.

How long does it take the robot to come momentarily to rest?

Solution 2

The robot comes to rest after \(\boxed{4.4\ \text{s}}\).

Choose right as positive. Then:

\[
u=+11\ \text{m s}^{-1}, \qquad a=-2.5\ \text{m s}^{-2}
\]

“Comes to rest” means the final velocity at that instant is:

\[
v=0
\]

Using:

\[
v=u+at
\]

we get:

\[
0=11+(-2.5)t
\]

\[
2.5t=11
\]

\[
t=\frac{11}{2.5}=4.4\ \text{s}
\]

The important point is that the robot’s acceleration is not zero when it stops. Its velocity is zero for an instant, but the leftward acceleration remains, so it would then begin moving left if the acceleration continued.

Question 3

An object has an initial velocity of \(-7.0\ \text{m s}^{-1}\) and a constant acceleration of \(-3.0\ \text{m s}^{-2}\).

A student says, “The negative acceleration means the object must be slowing down.”

Determine the object’s velocity after \(2.0\ \text{s}\), and assess the student’s claim.

Solution 3

The final velocity is \(\boxed{-13\ \text{m s}^{-1}}\), and the student is incorrect because the object is speeding up.

Using:

\[
v=u+at
\]

gives:

\[
v=-7.0+(-3.0)(2.0)
\]

\[
v=-7.0-6.0=-13\ \text{m s}^{-1}
\]

The initial speed was \(7.0\ \text{m s}^{-1}\). The final speed is \(13\ \text{m s}^{-1}\).

Both velocity and acceleration are negative, so they point in the same direction. The acceleration therefore increases the magnitude of the velocity.

The tempting misconception is to read a negative acceleration as “deceleration”. A negative sign only specifies direction relative to the chosen coordinate system.

Question 4

A cart moves to the right at \(8.0\ \text{m s}^{-1}\). A constant acceleration acts to the left. After \(3.0\ \text{s}\), the cart is moving to the left at \(1.0\ \text{m s}^{-1}\).

Find the cart’s acceleration. Then determine when, during the \(3.0\ \text{s}\) interval, it changed direction.

Solution 4

The cart’s acceleration is \(\boxed{-3.0\ \text{m s}^{-2}}\), and it changes direction after \(\boxed{2.7\ \text{s}}\) approximately.

Choose right as positive:

\[
u=+8.0\ \text{m s}^{-1}, \qquad v=-1.0\ \text{m s}^{-1}, \qquad t=3.0\ \text{s}
\]

First find the acceleration:

\[
a=\frac{v-u}{t}
\]

\[
a=\frac{-1.0-8.0}{3.0}
=\frac{-9.0}{3.0}
=-3.0\ \text{m s}^{-2}
\]

The negative sign means the acceleration points left.

To find when the cart changes direction, notice that it must first pass through zero velocity. Set \(v=0\):

\[
0=8.0+(-3.0)t
\]

\[
3.0t=8.0
\]

\[
t=\frac{8.0}{3.0}=2.67\ \text{s}
\]

So the cart comes momentarily to rest at about:

\[
\boxed{2.7\ \text{s}}
\]

It then spends the remaining \(0.33\ \text{s}\) moving left.

The hidden point is that a final negative velocity tells you more than “the object slowed down”. It proves that the object passed through \(v=0\) and reversed direction, provided the velocity changed continuously.

Question 5

Two identical carts begin with the same velocity of \(6.0\ \text{m s}^{-1}\) to the right.

Cart A has a constant acceleration of \(+2.0\ \text{m s}^{-2}\) for \(4.0\ \text{s}\).

Cart B has an acceleration of \(+4.0\ \text{m s}^{-2}\) for the first \(2.0\ \text{s}\), followed by \(-1.0\ \text{m s}^{-2}\) for the next \(2.0\ \text{s}\).

A student applies \(v=u+at\) to both carts using \(t=4.0\ \text{s}\). Explain why that approach works directly for Cart A but not for Cart B. Then calculate the final velocity of each cart.

Solution 5

Cart A finishes at \(\boxed{14\ \text{m s}^{-1}}\), while Cart B finishes at \(\boxed{12\ \text{m s}^{-1}}\). A single use of \(v=u+at\) works directly for Cart A because its acceleration is constant across the full \(4.0\ \text{s}\), but Cart B’s acceleration changes.

For Cart A:

\[
u=6.0\ \text{m s}^{-1}, \qquad a=2.0\ \text{m s}^{-2}, \qquad t=4.0\ \text{s}
\]

so:

\[
v=u+at
\]

\[
v=6.0+(2.0)(4.0)
=14\ \text{m s}^{-1}
\]

For Cart B, calculate each constant-acceleration interval separately.

During the first \(2.0\ \text{s}\):

\[
v_1=6.0+(4.0)(2.0)
=14\ \text{m s}^{-1}
\]

That becomes the initial velocity for the second interval.

During the next \(2.0\ \text{s}\):

\[
v_2=14+(-1.0)(2.0)
=12\ \text{m s}^{-1}
\]

Therefore:

\[
\boxed{v_A=14\ \text{m s}^{-1}}, \qquad
\boxed{v_B=12\ \text{m s}^{-1}}
\]

The trap is to treat acceleration as though it were one fixed number whenever an equation contains the symbol \(a\). The equation \(v=u+at\) assumes that the chosen \(a\) remains constant throughout the interval.

For Cart B, the two acceleration stages produce velocity changes of:

\[
\Delta v_1=(4.0)(2.0)=8.0\ \text{m s}^{-1}
\]

and:

\[
\Delta v_2=(-1.0)(2.0)=-2.0\ \text{m s}^{-1}
\]

giving a total velocity change of:

\[
\Delta v=8.0-2.0=6.0\ \text{m s}^{-1}
\]

That same reasoning leads naturally to acceleration-time graphs, where velocity change is found from the total signed area under the graph rather than from one rectangle \(at\).