When Is Average Velocity Half the Initial Plus Final Velocity?

A car travels east, increasing its velocity from \(+4.0\ \mathrm{m\,s^{-1}}\) to \(+16.0\ \mathrm{m\,s^{-1}}\) over \(3.0\ \mathrm{s}\). Is its average velocity \(+10.0\ \mathrm{m\,s^{-1}}\)?

Yes, provided the car has uniform acceleration throughout the entire interval.

For motion with constant acceleration, average velocity is the arithmetic mean of the initial and final velocities:

\[ \boxed{\bar v=\frac{u+v}{2}} \]

But there’s an important catch. Knowing the initial velocity, final velocity, and elapsed time isn’t enough to guarantee this result if acceleration changes.

For example, imagine the car reaches \(16\ \mathrm{m\,s^{-1}}\) in the first half-second, then maintains that velocity for the remaining \(2.5\ \mathrm{s}\). Its average velocity will be different, even though its initial velocity, final velocity, and total travel time are unchanged.

Let’s work out why.

01Why does the formula work?

Uniform acceleration means velocity changes by equal amounts in equal time intervals.

For example, a car accelerating uniformly from \(4\ \mathrm{m\,s^{-1}}\) to \(16\ \mathrm{m\,s^{-1}}\) over three seconds gains \(4\ \mathrm{m\,s^{-1}}\) every second.

Its velocity changes like this:

The velocity increases steadily. It doesn’t suddenly jump to its final value or spend most of the interval travelling at one particular velocity.

This produces a straight-line velocity-time graph.

The important observation is that the area under a velocity-time graph represents displacement.

For uniform acceleration, the area is a trapezium. Its parallel sides are the initial and final velocities, and its width is the elapsed time.

The trapezium area gives:

\[ s=\frac{1}{2}(u+v)t \]

Average velocity is displacement divided by elapsed time:

\[ \bar v=\frac{s}{t} \]

Substituting the trapezium area:

\[ \boxed{\bar v=\frac{u+v}{2}} \]

This is why the formula works. It isn’t an arbitrary shortcut. It follows directly from the straight-line relationship between velocity and time during uniform acceleration.

What do the symbols mean?

Notice that \(u\) and \(v\) represent velocities, not speeds. Their signs matter.

02Signs matter: what if the object changes direction?

An object can have positive velocity at the beginning of a journey and negative velocity at the end.

Does this mean the average-velocity formula stops working?

No. It still works under uniform acceleration, provided you use signed velocities.

Imagine a cyclist travelling east at \(6\ \mathrm{m\,s^{-1}}\). The cyclist experiences a constant acceleration of \(-2\ \mathrm{m\,s^{-2}}\) for five seconds.

Take east as positive.

The final velocity is:

\[ v=u+at \]

\[ v=6+(-2)(5)=-4\ \mathrm{m\,s^{-1}} \]

The cyclist finishes travelling west.

The average velocity is:

\[ \bar v=\frac{6+(-4)}{2} \]

\[ \boxed{\bar v=+1\ \mathrm{m\,s^{-1}}} \]

The positive answer means the net displacement is eastwards.

But notice something interesting: the cyclist actually changes direction during the journey.

The velocity reaches zero after three seconds:

\[ 0=6-2t \]

\[ t=3\ \mathrm{s} \]

During the first three seconds, the cyclist travels \(9\ \mathrm{m}\) east. During the remaining two seconds, the cyclist travels \(4\ \mathrm{m}\) west.

Therefore:

• Net displacement = \(9-4=5\ \mathrm{m}\) east.
• Total distance = \(9+4=13\ \mathrm{m}\).

The average velocity is:

\[ \bar v=\frac{5}{5}=1\ \mathrm{m\,s^{-1}} \]

The average speed, however, is:

\[ \text{Average speed}=\frac{13}{5} =2.6\ \mathrm{m\,s^{-1}} \]

This gives us an important distinction:

• Average velocity uses net displacement.
• Average speed uses total distance travelled.

The formula \((u+v)/2\) gives average velocity, not average speed.

A negative acceleration also doesn’t automatically mean an object is slowing down. Here, the cyclist initially slows down while travelling east, stops momentarily, and then speeds up while travelling west.

For more on this distinction, see How to Choose Positive Direction in HSC Vertical Motion.

03When does the formula fail?

The formula is guaranteed to work when acceleration is constant throughout the interval.

If acceleration changes, the velocity-time graph is no longer one straight line. The area underneath it might not equal the trapezium formed by joining the initial and final velocities.

Let’s return to our car.

It starts at \(4\ \mathrm{m\,s^{-1}}\) and finishes at \(16\ \mathrm{m\,s^{-1}}\), taking three seconds in total.

Consider two possible journeys.

Journey A: Uniform acceleration

The car increases its velocity steadily from \(4\) to \(16\ \mathrm{m\,s^{-1}}\) over three seconds.

Its average velocity is:

\[ \bar v=\frac{4+16}{2} =10\ \mathrm{m\,s^{-1}} \]

Its displacement is \(10\times3=30\ \mathrm{m}\).

Journey B: Changing acceleration

The car accelerates from \(4\) to \(16\ \mathrm{m\,s^{-1}}\) in just \(0.5\) seconds, then maintains \(16\ \mathrm{m\,s^{-1}}\) for the remaining \(2.5\) seconds.

Its displacement during the acceleration phase is:

\[ s_1=\frac{4+16}{2}\times0.5 =5\ \mathrm{m} \]

Its displacement during the constant-velocity phase is:

\[ s_2=16\times2.5 =40\ \mathrm{m} \]

Total displacement:

\[ s=5+40=45\ \mathrm{m} \]

Therefore:

\[ \bar v=\frac{45}{3} =15\ \mathrm{m\,s^{-1}} \]

The comparison is clear:

Both journeys have identical initial and final velocities. However, their average velocities are different.

The reason is that the car in Journey B spends much more time travelling close to its final velocity.

An important qualification

Uniform acceleration is a sufficient condition for using the formula. It is not strictly the only situation in which the formula can give the correct numerical answer.

A changing-acceleration motion can sometimes produce exactly the same total area under its velocity-time graph as the straight-line trapezium joining its endpoints.

In that situation, the formula happens to give the correct average velocity, even though acceleration isn’t constant.

However, you cannot establish this just by looking at the initial and final velocities. You need additional information, such as the displacement or the complete velocity-time graph.

This distinction becomes important in more challenging HSC questions.

04A quick decision process

Before using \(\bar v=(u+v)/2\), check the following:

Let’s apply these ideas to some HSC-style questions.

05Worked examples

06Questions and solutions

Question 1

A trolley moves east along a straight track with uniform acceleration. Its initial velocity is \(+5.0\ \mathrm{m\,s^{-1}}\), and its final velocity is \(+13.0\ \mathrm{m\,s^{-1}}\) after \(4.0\ \mathrm{s}\).

Calculate:

a) Its average velocity.

b) Its displacement during the four seconds.

Solution 1

The average velocity is \(+9.0\ \mathrm{m\,s^{-1}}\), and the displacement is \(+36\ \mathrm{m}\).

Step 1: Apply the average-velocity equation.

Because acceleration is uniform:

\[ \begin{aligned} \bar v&=\frac{u+v}{2}\\ &=\frac{5.0+13.0}{2}\\ &=+9.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 2: Calculate displacement.

\[ \begin{aligned} s&=\bar vt\\ &=9.0\times4.0\\ &=+36\ \mathrm{m} \end{aligned} \]

The positive sign indicates displacement eastwards.

The important step was recognising that acceleration is uniform. Without that information, the initial and final velocities alone wouldn’t establish the average velocity.

Question 2

A cyclist travels east at \(12\ \mathrm{m\,s^{-1}}\). The cyclist experiences a constant acceleration of \(-2.0\ \mathrm{m\,s^{-2}}\) for \(10\ \mathrm{s}\).

Take east as positive.

Calculate:

a) The final velocity.

b) The average velocity.

c) The displacement.

d) The average speed.

Solution 2

The final velocity is \(-8.0\ \mathrm{m\,s^{-1}}\), the average velocity is \(+2.0\ \mathrm{m\,s^{-1}}\), the displacement is \(+20\ \mathrm{m}\), and the average speed is \(5.2\ \mathrm{m\,s^{-1}}\).

Step 1: Calculate the final velocity.

Use the uniform-acceleration equation:

\[ \begin{aligned} v&=u+at\\ &=12+(-2.0)(10)\\ &=-8.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

The negative sign means the cyclist finishes travelling west.

Step 2: Calculate average velocity.

Acceleration is constant over the entire interval, so:

\[ \begin{aligned} \bar v&=\frac{u+v}{2}\\ &=\frac{12+(-8)}{2}\\ &=+2.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 3: Calculate displacement.

\[ \begin{aligned} s&=\bar vt\\ &=2.0\times10\\ &=+20\ \mathrm{m} \end{aligned} \]

The cyclist finishes \(20\ \mathrm{m}\) east of the starting position.

Step 4: Calculate average speed.

We cannot use average velocity to calculate average speed directly because the cyclist reverses direction.

First, find when the cyclist stops momentarily:

\[ \begin{aligned} 0&=12-2t\\ t&=6.0\ \mathrm{s} \end{aligned} \]

The cyclist travels east for six seconds, then west for four seconds.

During the first six seconds:

\[ s_1=\frac{12+0}{2}\times6 =36\ \mathrm{m} \]

During the final four seconds:

\[ s_2=\frac{0+(-8)}{2}\times4 =-16\ \mathrm{m} \]

The total distance is:

\[ d=36+16=52\ \mathrm{m} \]

Therefore:

\[ \text{Average speed}=\frac{52}{10} \]

\[ \boxed{5.2\ \mathrm{m\,s^{-1}}} \]

The tempting mistake is to use the magnitude of the average velocity as the average speed. That would give \(2.0\ \mathrm{m\,s^{-1}}\), which ignores the distance travelled after the cyclist changes direction.

Question 3

An electric cart moves along a straight track. A motion sensor records the following velocities.

Assume the velocity changes linearly between consecutive readings.

Calculate the average velocity over the six-second interval.

A student suggests using the initial and final velocities in the equation \(\bar v=(u+v)/2\). Explain whether this is justified.

Solution 3

The average velocity is approximately \(+9.67\ \mathrm{m\,s^{-1}}\). The student’s endpoint shortcut is not justified because acceleration is not constant throughout the six seconds.

Step 1: Calculate the displacement during each interval.

The velocity changes linearly within each two-second interval, so acceleration is constant within each individual section.

For the first two seconds:

\[ s_1=\frac{2+8}{2}\times2 =10\ \mathrm{m} \]

For the next two seconds:

\[ s_2=\frac{8+10}{2}\times2 =18\ \mathrm{m} \]

For the final two seconds:

\[ s_3=\frac{10+20}{2}\times2 =30\ \mathrm{m} \]

Step 2: Calculate total displacement.

\[ s=10+18+30 \]

\[ s=58\ \mathrm{m} \]

Step 3: Calculate average velocity.

\[ \begin{aligned} \bar v&=\frac{s}{t}\\ &=\frac{58}{6}\\ &=+9.67\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 4: Evaluate the student’s method.

The acceleration in each interval is:

\[ a_1=\frac{8-2}{2}=3\ \mathrm{m\,s^{-2}} \]

\[ a_2=\frac{10-8}{2}=1\ \mathrm{m\,s^{-2}} \]

\[ a_3=\frac{20-10}{2}=5\ \mathrm{m\,s^{-2}} \]

These accelerations are different.

Therefore, acceleration is not uniform over the entire six-second interval.

The student’s calculation would give:

\[ \frac{2+20}{2}=11\ \mathrm{m\,s^{-1}} \]

This would predict a displacement of \(66\ \mathrm{m}\), rather than the actual \(58\ \mathrm{m}\).

The key is to use the area under the complete velocity-time graph when acceleration changes.

Question 4

A train travels east along a straight track for eight seconds.

Its motion consists of two stages:

• During the first \(5.0\ \mathrm{s}\), it accelerates uniformly from \(+2.0\) to \(+12.0\ \mathrm{m\,s^{-1}}\).
• During the next \(3.0\ \mathrm{s}\), it accelerates uniformly from \(+12.0\) to \(+6.0\ \mathrm{m\,s^{-1}}\).

Calculate the average velocity over the entire eight-second journey.

Explain why using the initial and final velocities directly gives an incorrect answer. Also explain why simply averaging the two stage-average velocities is not quite correct.

Solution 4

The overall average velocity is \(+7.75\ \mathrm{m\,s^{-1}}\).

The correct method is to calculate displacement during each stage and divide the combined displacement by the total time.

Step 1: Calculate the average velocity during the first stage.

Acceleration is uniform during the first five seconds:

\[ \bar v_1=\frac{2+12}{2} =7.0\ \mathrm{m\,s^{-1}} \]

The displacement is:

\[ s_1=7.0\times5 =35\ \mathrm{m} \]

Step 2: Calculate the average velocity during the second stage.

Again, acceleration is uniform within this stage:

\[ \bar v_2=\frac{12+6}{2} =9.0\ \mathrm{m\,s^{-1}} \]

The displacement is:

\[ s_2=9.0\times3 =27\ \mathrm{m} \]

Step 3: Calculate the overall average velocity.

Total displacement:

\[ s=35+27=62\ \mathrm{m} \]

Total time:

\[ t=5+3=8\ \mathrm{s} \]

Therefore:

\[ \begin{aligned} \bar v&=\frac{62}{8}\\ &=+7.75\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 4: Understand the two tempting mistakes.

The first mistake is applying the endpoint formula to the whole journey:

\[ \frac{2+6}{2}=4.0\ \mathrm{m\,s^{-1}} \]

This is incorrect because acceleration changes between the two stages.

The second mistake is taking the ordinary arithmetic mean of the two stage-average velocities:

\[ \frac{7+9}{2}=8.0\ \mathrm{m\,s^{-1}} \]

This is also incorrect because the stages have different durations.

The first average velocity applies for five seconds, while the second applies for only three seconds.

The correct overall average is a time-weighted average:

\[ \bar v=\frac{7(5)+9(3)}{5+3} \]

\[ \boxed{\bar v=7.75\ \mathrm{m\,s^{-1}}} \]

Whenever you combine averages from different time intervals, the duration of each interval matters.

Question 5

This final question tests a more subtle distinction: can the endpoint formula give the correct average velocity even when acceleration is not uniform?

A cart moves east along a straight track for \(6.0\ \mathrm{s}\). A motion logger records the following velocities:

An independent position sensor measures the cart’s total displacement as \(+48\ \mathrm{m}\).

Assume that velocity changes linearly between each pair of consecutive readings, with no sudden jumps in velocity.

a) Determine the missing velocity reading at \(t=4.0\ \mathrm{s}\).

b) Calculate the acceleration during each two-second interval. Is acceleration uniform over the entire six seconds?

c) Determine the actual average velocity.

d) A student notices that the actual average velocity equals half the initial velocity plus the final velocity. The student concludes that acceleration must therefore have been uniform. Explain why this conclusion is incorrect.

e) If the assumption of linear velocity changes between readings were removed, could the missing velocity be determined uniquely from the remaining information? Explain.

Solution 5

The missing velocity is \(+4.0\ \mathrm{m\,s^{-1}}\). Acceleration is not uniform, yet the endpoint-average formula gives the correct average velocity of \(+8.0\ \mathrm{m\,s^{-1}}\) for this particular journey.

This is possible because the areas above and below the straight-line velocity graph joining the endpoints cancel out.

Step 1: Find the missing velocity.

Let the missing velocity at \(t=4.0\ \mathrm{s}\) be \(w\).

The velocity changes linearly between readings, so each two-second section has uniform acceleration. We can therefore calculate displacement in each section using the endpoint-average formula.

For the first interval:

\[ s_1=\frac{2+12}{2}\times2=14\ \mathrm{m} \]

For the second interval:

\[ s_2=\frac{12+w}{2}\times2 =12+w \]

For the third interval:

\[ s_3=\frac{w+14}{2}\times2 =w+14 \]

The total displacement is \(48\ \mathrm{m}\):

\[ 48=14+(12+w)+(w+14) \]

Simplifying:

\[ 48=40+2w \]

\[ 2w=8 \]

\[ \boxed{w=+4.0\ \mathrm{m\,s^{-1}}} \]

Step 2: Calculate the acceleration in each interval.

Use the change in velocity divided by the time interval.

First interval:

\[ a_1=\frac{12-2}{2} =+5.0\ \mathrm{m\,s^{-2}} \]

Second interval:

\[ a_2=\frac{4-12}{2} =-4.0\ \mathrm{m\,s^{-2}} \]

Third interval:

\[ a_3=\frac{14-4}{2} =+5.0\ \mathrm{m\,s^{-2}} \]

Therefore:

\[ \boxed{a_1\ne a_2\ne a_3} \]

Acceleration is clearly not uniform over the entire six seconds.

The cart accelerates east, then accelerates west, then accelerates east again.

Step 3: Calculate the actual average velocity.

We already know the total displacement and time:

\[ \begin{aligned} \bar v&=\frac{s}{t}\\ &=\frac{48}{6}\\ &=+8.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Now compare this with the endpoint formula:

\[ \begin{aligned} \frac{u+v}{2} &=\frac{2+14}{2}\\ &=+8.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Both methods give the same result.

Step 4: Explain why the formula works in this particular case.

This is the subtle part.

If acceleration were uniform from \(+2\) to \(+14\ \mathrm{m\,s^{-1}}\) over six seconds, the velocity would increase by \(4\ \mathrm{m\,s^{-1}}\) every two seconds.

The corresponding velocities would be:

The actual velocity is \(6\ \mathrm{m\,s^{-1}}\) higher than the uniform-acceleration prediction at two seconds.

At four seconds, it is \(6\ \mathrm{m\,s^{-1}}\) lower.

On a velocity-time graph, the actual graph therefore creates extra positive area early in the journey and an equal amount of negative area later.

The two differences cancel.

Consequently, the actual displacement is exactly the same as the displacement predicted by the straight-line graph joining the initial and final velocities.

The endpoint formula gives the correct numerical average, even though acceleration changes.

Step 5: Consider what happens without the linear-change assumption.

The missing velocity could no longer be determined uniquely.

The displacement measurement tells us the total signed area under the velocity-time graph. It does not tell us the exact shape of that graph between readings.

Many different velocity histories could have the same initial velocity, final velocity, recorded intermediate velocity, and total displacement.

The assumption of linear changes between readings is what allowed us to calculate the missing velocity.

This is the distinction to remember:

• Uniform acceleration guarantees that \(\bar v=(u+v)/2\).
• Nonuniform acceleration means the formula is not guaranteed.
• However, nonuniform acceleration does not make the formula mathematically impossible. It can still give the correct answer if the actual signed area matches the endpoint trapezium.

07The key decision in an HSC question

When you see initial and final velocities, don’t automatically reach for the average-velocity shortcut.

First establish whether acceleration is uniform over the interval you’re considering. If it is, use the formula confidently, remembering to keep the velocity signs.

If acceleration changes, think about the velocity-time graph and the displacement represented by its signed area. If the motion has multiple stages, treat each stage separately before combining the results.

For projectile questions, the next useful step is often resolving the initial velocity into horizontal and vertical components before finding the final velocity. See Final Velocity of a Projectile: HSC Physics Guide for that related calculation.