Why Greater Drag Requires More Power to Maintain Vehicle Speed

Two identical cars are travelling along a level road at 25 m/s. One has a clean roofline. The other carries a large roof box. Both drivers maintain exactly the same speed.

Which car needs more power from its engine?

The car with the roof box needs more power, assuming the box increases aerodynamic drag. Its engine must provide a greater forward force to balance the stronger backward force from the air.

Here’s the interesting part: neither car is speeding up. Their kinetic energies remain constant, yet one engine must transfer energy at a greater rate. Understanding why requires separating two quantities that students often mix up: work and power.

01Why constant speed does not mean constant power demand

Imagine pushing a shopping trolley across two different surfaces.

On smooth tiles, you need relatively little force to keep it moving at a steady walking speed. On thick carpet, you need to push harder to maintain that same speed.

The trolley isn’t accelerating in either situation. Your forward push simply balances the resistance opposing its motion.

A vehicle works in a similar way. Its engine supplies a driving force through the wheels, while several resistive forces act against its motion.

For a vehicle travelling along a level road, the main horizontal forces are:

• Driving force: the forward force supplied through the tyres.
• Aerodynamic drag: the backward force caused by the vehicle moving through air.
• Rolling resistance: resistance associated with the tyres and their interaction with the road.

Choose forward motion as the positive direction. Newton’s second law gives:

\[ F_{\mathrm{drive}}-F_{\mathrm{drag}} -F_{\mathrm{rolling}}=ma \]

When the vehicle maintains constant speed, its acceleration is zero.

Therefore:

\[ F_{\mathrm{drive}}= F_{\mathrm{drag}}+F_{\mathrm{rolling}} \]

This is the important relationship.

If aerodynamic drag increases, the driving force must also increase to maintain constant speed.

The trolley analogy helps explain the balance of forces, but it has a limitation: carpet resistance doesn’t behave exactly like aerodynamic drag. Air resistance also depends on the vehicle’s speed relative to the surrounding air, its shape, and its frontal area.

We need a more precise model to connect this increased force to power.

02From work to power: why greater drag requires more energy per second

Work measures the amount of energy transferred when a force acts through a displacement.

For a constant force acting in the direction of motion:

\[ W=Fd \]

where:

• \(W\) is work done, measured in joules (J).
• \(F\) is force, measured in newtons (N).
• \(d\) is displacement, measured in metres (m).

Power measures how quickly that work is done:

\[ P=\frac{W}{t} \]

Substituting \(W=Fd\):

\[ P=\frac{Fd}{t} \]

Since speed is displacement divided by time:

\[ \boxed{P=Fv} \]

Here, \(v\) is the vehicle’s speed in metres per second.

This equation gives us the direct answer to the original question.

For a vehicle maintaining the same speed:

\[ P_{\mathrm{drag}}=F_{\mathrm{drag}}v \]

Therefore, increasing the drag force increases the power required to overcome it.

If the drag force doubles while speed remains unchanged, the power required to overcome that drag also doubles.

Work and power are not interchangeable

This distinction is worth getting comfortable with.

One watt is one joule transferred per second.

Consider two vehicles travelling at the same speed. One experiences 300 N of aerodynamic drag, while the other experiences 450 N.

Both travel 20 metres every second.

The second vehicle experiences a greater opposing force over the same distance each second. Consequently, its engine must transfer more energy every second.

That is greater power.

There is another important detail here. If the vehicle maintains constant speed, its kinetic energy does not increase.

The engine does positive work, while aerodynamic drag and rolling resistance do negative work. These contributions balance, leaving the vehicle’s kinetic energy unchanged.

So, constant speed does not mean zero work or zero engine power. It means zero net work on the vehicle’s kinetic energy.

For a related explanation, see Can an Object Have Kinetic Energy While Net Power Is Zero?.

03How can we experimentally measure the effect of greater drag?

So far, we’ve used equations to predict what should happen.

But how could we test whether adding a roof box actually increases the resistive force acting on a vehicle?

One useful method is a coast-down investigation.

The idea is straightforward: allow a vehicle to slow down without engine power driving it, then measure how quickly its speed decreases.

Setting up the investigation

Consider a vehicle travelling along a straight, level test track.

A radar speed sensor is positioned beside the track and connected to a data logger. The sensor records the vehicle’s speed at short time intervals.

The investigation has two configurations:

1. The vehicle travels without a roof box.
2. The same vehicle travels with a roof box installed.

For both tests, the transmission is placed in neutral, the brakes are released, and the vehicle is allowed to coast.

The total vehicle mass is kept equal in both configurations using appropriate ballast. Tyre pressures, tyre condition, and other relevant conditions are also kept as consistent as possible.

The data logger records speed against time.

What should the experiment show?

Predict what will happen when the roof box is installed.

If it increases aerodynamic drag, the vehicle should experience a greater opposing force at the same speed.

With no engine driving force, that greater resistance should produce a greater deceleration.

In other words, the roof-box configuration should show a more negative acceleration.

Understanding the measurements

Suppose the following illustrative data are obtained.

The vehicle has a mass of 1500 kg in both configurations. Its acceleration is measured when its speed is 25 m/s.

The negative signs indicate that the vehicle is slowing down.

Newton’s second law allows us to calculate the magnitude of the total resistive force.

\[ F_{\mathrm{resistive}}=m|a| \]

Without the roof box:

\[ F_1=1500\times0.20 \]

\[ F_1=300\text{ N} \]

With the roof box:

\[ F_2=1500\times0.32 \]

\[ F_2=480\text{ N} \]

The additional resistive force is:

\[ \Delta F=480-300 \]

\[ \boxed{\Delta F=180\text{ N}} \]

Assuming other resistive forces remain unchanged, the roof box has increased the resistance by approximately 180 N at 25 m/s.

How much additional power would the engine need to overcome this resistance at the same speed?

\[ \Delta P=\Delta Fv \]

\[ \Delta P=180\times25 \]

\[ \boxed{\Delta P=4500\text{ W}=4.5\text{ kW}} \]

The vehicle therefore needs an additional 4.5 kW of wheel power to maintain 25 m/s under these conditions.

Why the experimental design matters

A simple comparison of total coast-down times can be misleading.

Aerodynamic drag changes as the vehicle slows down. Comparing runs over different speed ranges would mix together different drag forces.

A better approach is to compare the instantaneous acceleration at the same speed, using the gradient of the speed-time graph.

The investigation should also be repeated in both directions along the track to reduce the influence of wind and small road gradients.

There is one limitation worth recognising: a coast-down test measures the combined effect of aerodynamic drag, rolling resistance, and mechanical resistance. It does not automatically isolate aerodynamic drag alone.

The comparison works because these other contributions are assumed to remain approximately unchanged between matched tests.

This is an example of how experimental controls allow us to identify the effect of one changed variable.

04Why aerodynamic drag becomes especially important at higher speeds

So far, we’ve considered situations where the vehicle maintains the same speed but experiences different drag forces.

However, aerodynamic drag also changes when the vehicle’s speed changes.

A useful model for aerodynamic drag is:

\[ F_D=\frac{1}{2}\rho C_D A v_{\mathrm{air}}^2 \]

where:

• \(F_D\) is aerodynamic drag force in newtons.
• \(\rho\) is the density of air in kg/m³.
• \(C_D\) is the drag coefficient, which describes the vehicle’s aerodynamic shape.
• \(A\) is the vehicle’s frontal area in m².
• \(v_{\mathrm{air}}\) is the vehicle’s speed relative to the surrounding air in m/s.

For the same vehicle, in the same air density and under comparable conditions:

\[ F_D\propto v_{\mathrm{air}}^2 \]

This means that doubling the relative air speed produces four times the aerodynamic drag force.

But remember our power equation:

\[ P_D=F_Dv \]

In still air, the vehicle’s speed relative to the air equals its speed relative to the road.

Therefore:

\[ P_D\propto v^2\times v \]

\[ \boxed{P_D\propto v^3} \]

Aerodynamic power demand increases approximately with the cube of speed under these assumptions.

Consider the following illustrative data for a vehicle in still air.

Notice what happens when speed increases from 10 m/s to 20 m/s.

• Speed doubles.
• Drag force increases by a factor of four.
• Power against aerodynamic drag increases by a factor of eight.

This is why increasing a vehicle’s cruising speed can require a substantial increase in power.

Be careful with the conclusion, though. The cubic relationship applies to aerodynamic power under the stated assumptions. Total wheel power also includes rolling resistance and other loads, so it does not necessarily increase by exactly eight times when speed doubles.

There is another subtlety: the speed used to calculate aerodynamic drag is not always the same as the speed used to calculate wheel power.

If a vehicle travels into a headwind, its speed relative to the air is greater than its speed relative to the road.

We will return to this distinction in the final question.

05Wheel power, engine power, and efficiency

There is one more distinction to make before tackling exam-style questions.

When we calculate \(P=Fv\) using the total resistive force, we find the mechanical power that must be delivered at the wheels.

This is not necessarily the same as the chemical power supplied by the fuel.

An engine and drivetrain have energy losses. Some of the energy supplied by the fuel is transferred to the surroundings rather than reaching the wheels as useful mechanical power.

Efficiency describes how much of the input power becomes useful output power.

\[ \eta=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}} \]

For a vehicle, if we are given the overall fuel-to-wheel efficiency:

\[ \eta=\frac{P_{\mathrm{wheel}}}{P_{\mathrm{fuel}}} \]

Rearranging:

\[ \boxed{P_{\mathrm{fuel}}= \frac{P_{\mathrm{wheel}}}{\eta}} \]

Notice that we divide by efficiency, not multiply.

For example, if a vehicle has an overall fuel-to-wheel efficiency of 25%, only 25% of the chemical power supplied by the fuel reaches the wheels.

The fuel input power must therefore be four times the wheel output power.

Be precise about which efficiency a question gives you:

• Overall fuel-to-wheel efficiency compares chemical power from fuel with mechanical power at the wheels.
• Drivetrain efficiency compares mechanical power at the engine shaft with mechanical power reaching the wheels.

These are different quantities.

If a question gives drivetrain efficiency only, you should use it to convert between engine shaft power and wheel power. Do not automatically treat it as overall fuel efficiency.

06A reliable method for vehicle power questions

When you encounter a question about a vehicle maintaining constant speed, work through these decisions in order.

1. Identify the forces. Separate aerodynamic drag from rolling resistance and any other opposing forces.

2. Check whether speed is constant. If acceleration is zero, the total driving force balances the total opposing force on a level road.

3. Calculate the wheel power. Use the driving force and vehicle speed:

\[ P_{\mathrm{wheel}}=F_{\mathrm{drive}}v \]

1. Check whether efficiency is involved. If the question asks for fuel input power, use the stated efficiency with the correct input and output convention.

2. Check whether the question asks for energy or power. If it asks how much energy is transferred over a period of time:

\[ E=Pt \]

One particularly common mistake is using the net force in the wheel-power equation.

If a vehicle travels at constant speed, its net force is zero. That does not mean its engine delivers zero power.

The engine’s driving force balances the resistive forces, but the engine still transfers energy at a positive rate.

For a more detailed treatment of energy transferred against resistance, see Work Against Resistive Forces in HSC Physics.

07Questions and solutions

Question 1

A car travels at a constant speed of 22 m/s along a level road. The total resistive force acting on it is 420 N.

(a) Calculate the mechanical power that must be delivered at the wheels.

(b) Calculate the energy supplied at the wheels over a period of 3.0 minutes.

Solution 1

The wheels must deliver 9.24 kW of power, supplying approximately 1.66 MJ of energy over 3.0 minutes.

Part (a)

At constant speed, the driving force balances the total resistive force.

Therefore:

\[ F_{\mathrm{drive}}=420\text{ N} \]

Using:

\[ P=Fv \]

Substitute:

\[ P=420\times22 \]

\[ \boxed{P=9240\text{ W}=9.24\text{ kW}} \]

Part (b)

First, convert the time into seconds:

\[ t=3.0\times60=180\text{ s} \]

The energy transferred is:

\[ E=Pt \]

Substitute:

\[ E=9240\times180 \]

\[ E=1663200\text{ J} \]

Therefore:

\[ \boxed{E=1.66\text{ MJ}} \]

The power describes the rate of energy transfer, while the energy describes the total amount transferred over the three minutes.

The vehicle’s kinetic energy remains constant because its speed does not change.

Question 2

A vehicle travels along a level road in still air.

At a speed of 15 m/s, it experiences:

• Aerodynamic drag of 180 N.
• Rolling resistance of 220 N.

The vehicle then increases its constant speed to 30 m/s.

Assume:

• Aerodynamic drag is proportional to the square of speed.
• Rolling resistance remains constant.
• Air density and vehicle orientation do not change.

(a) Calculate the total wheel power at 15 m/s.

(b) Calculate the aerodynamic drag at 30 m/s.

(c) Calculate the total wheel power at 30 m/s.

(d) Determine the ratio of the two wheel powers and explain why it is not equal to eight.

Solution 2

The wheel power is 6.0 kW at 15 m/s and 28.2 kW at 30 m/s. The power increases by a factor of 4.7, rather than eight, because rolling resistance does not increase with the square of speed in this model.

Part (a)

First, calculate the total resistive force at 15 m/s.

\[ F_{\mathrm{total}}=180+220 \]

\[ F_{\mathrm{total}}=400\text{ N} \]

Now calculate wheel power:

\[ P_1=F_{\mathrm{total}}v \]

\[ P_1=400\times15 \]

\[ \boxed{P_1=6000\text{ W}=6.0\text{ kW}} \]

Part (b)

The speed has doubled.

Since aerodynamic drag is proportional to speed squared:

\[ F_{\mathrm{drag,2}}= F_{\mathrm{drag,1}} \left(\frac{v_2}{v_1}\right)^2 \]

Substitute:

\[ F_{\mathrm{drag,2}}= 180\left(\frac{30}{15}\right)^2 \]

\[ F_{\mathrm{drag,2}}=180\times4 \]

\[ \boxed{F_{\mathrm{drag,2}}=720\text{ N}} \]

Part (c)

Rolling resistance remains 220 N.

Therefore:

\[ F_{\mathrm{total,2}}=720+220 \]

\[ F_{\mathrm{total,2}}=940\text{ N} \]

Calculate the new wheel power:

\[ P_2=940\times30 \]

\[ \boxed{P_2=28200\text{ W}=28.2\text{ kW}} \]

Part (d)

Calculate the power ratio:

\[ \frac{P_2}{P_1}= \frac{28.2}{6.0} \]

\[ \boxed{\frac{P_2}{P_1}=4.7} \]

Why isn’t the ratio eight?

If aerodynamic drag were the only resistance, doubling speed would increase drag fourfold and power eightfold.

However, rolling resistance also contributes to wheel power. In this question, it remains constant, so its power contribution only doubles when speed doubles.

The total power therefore increases by a factor of 4.7.

The important lesson is to apply the speed-squared relationship to aerodynamic drag, not automatically to every resistive force.

Question 3

A vehicle undergoes a coast-down investigation on a level test track.

Its mass is maintained at 1500 kg in both configurations.

A radar speed sensor records the vehicle’s acceleration when it is travelling at 25 m/s.

The following data are obtained:

Assume that other resistive forces remain unchanged between the two tests.

(a) Calculate the total resistive force in each configuration.

(b) Determine the additional resistive force associated with the roof box.

(c) Calculate the additional wheel power needed to maintain 25 m/s with the roof box installed.

(d) Explain why the acceleration measurements must be compared at the same speed. Identify one experimental control and one limitation of the investigation.

Solution 3

The roof box introduces an additional resistive force of 180 N. Maintaining 25 m/s therefore requires an additional 4.5 kW of wheel power.

Part (a)

Since the vehicle is coasting, the net horizontal force is the resultant resistive force.

Using Newton’s second law:

\[ F_{\mathrm{net}}=ma \]

We use the magnitude of acceleration to calculate the magnitude of resistance.

Without the roof box:

\[ F_1=1500\times0.20 \]

\[ \boxed{F_1=300\text{ N}} \]

With the roof box:

\[ F_2=1500\times0.32 \]

\[ \boxed{F_2=480\text{ N}} \]

Part (b)

Subtract the two resistive forces:

\[ \Delta F=480-300 \]

\[ \boxed{\Delta F=180\text{ N}} \]

The roof-box configuration experiences 180 N of additional resistance.

This difference can be attributed to the roof box under the assumption that the other resistive forces remain unchanged.

Part (c)

The additional wheel power required is:

\[ \Delta P=\Delta Fv \]

Substitute:

\[ \Delta P=180\times25 \]

\[ \boxed{\Delta P=4500\text{ W}=4.5\text{ kW}} \]

Part (d)

The acceleration measurements must be compared at the same speed because aerodynamic drag depends on the vehicle’s speed relative to the air.

Comparing acceleration at different speeds could produce a difference caused by speed rather than the roof box.

One suitable experimental control is maintaining the same tyre pressure and vehicle mass in both configurations. Repeating trials in both directions would also help reduce the influence of wind and road gradients.

A limitation is that coast-down measurements include rolling resistance and mechanical resistance as well as aerodynamic drag. The difference between matched tests estimates the additional resistance only if those other contributions remain unchanged.

Question 4

A 1500 kg car travels at 24 m/s along a level road at constant speed.

Its wheel power is 18 kW, and its rolling resistance is 150 N.

A roof box is installed, increasing the aerodynamic drag force by 100 N.

The driver initially leaves the wheel power unchanged.

Assume that:

• The vehicle’s mass remains unchanged.
• Rolling resistance remains constant.
• The vehicle is initially travelling at constant speed.
• The overall fuel-to-wheel efficiency is 30% and remains constant.

(a) Calculate the initial driving force and initial aerodynamic drag.

(b) Immediately after the roof box is installed, calculate the net force and acceleration.

(c) Calculate the new wheel power required to maintain 24 m/s.

(d) Calculate the additional fuel input power required to maintain the original speed.

Solution 4

The initial driving force is 750 N, and the initial aerodynamic drag is 600 N. After the roof box is installed, the vehicle initially decelerates at approximately \(0.067\text{ m/s}^2\). Restoring its original speed requires 20.4 kW of wheel power and an additional 8.0 kW of fuel input power.

Part (a)

The initial wheel power is 18 kW.

Convert to watts:

\[ P_{\mathrm{wheel}}=18000\text{ W} \]

Using:

\[ P=Fv \]

Rearrange:

\[ F_{\mathrm{drive}}=\frac{P}{v} \]

Substitute:

\[ F_{\mathrm{drive}}=\frac{18000}{24} \]

\[ \boxed{F_{\mathrm{drive}}=750\text{ N}} \]

The vehicle initially travels at constant speed, so the total resistive force must also equal 750 N.

Therefore:

\[ F_{\mathrm{drag}}= F_{\mathrm{drive}}-F_{\mathrm{rolling}} \]

\[ F_{\mathrm{drag}}=750-150 \]

\[ \boxed{F_{\mathrm{drag}}=600\text{ N}} \]

Part (b)

The roof box adds 100 N of aerodynamic drag.

The new aerodynamic drag is:

\[ F_{\mathrm{drag,new}}=600+100 \]

\[ F_{\mathrm{drag,new}}=700\text{ N} \]

The total resistive force becomes:

\[ F_{\mathrm{resistive}}=700+150 \]

\[ F_{\mathrm{resistive}}=850\text{ N} \]

However, the driving force remains 750 N because the driver has not increased the wheel power.

Taking forward as positive:

\[ F_{\mathrm{net}}=750-850 \]

\[ F_{\mathrm{net}}=-100\text{ N} \]

Using Newton’s second law:

\[ a=\frac{F_{\mathrm{net}}}{m} \]

\[ a=\frac{-100}{1500} \]

\[ \boxed{a=-0.0667\text{ m/s}^2} \]

The negative acceleration means the vehicle is slowing down.

Part (c)

To restore constant speed, the driving force must again balance the total resistive force.

Therefore:

\[ F_{\mathrm{drive,new}}=850\text{ N} \]

Calculate the required wheel power:

\[ P_{\mathrm{wheel,new}}=850\times24 \]

\[ \boxed{P_{\mathrm{wheel,new}}=20.4\text{ kW}} \]

The additional wheel power is:

\[ \Delta P=20.4-18.0 \]

\[ \boxed{\Delta P=2.4\text{ kW}} \]

Part (d)

The overall fuel-to-wheel efficiency is:

\[ \eta=0.30 \]

The additional fuel input power is:

\[ \Delta P_{\mathrm{fuel}}= \frac{\Delta P_{\mathrm{wheel}}}{\eta} \]

Substitute:

\[ \Delta P_{\mathrm{fuel}}= \frac{2.4}{0.30} \]

\[ \boxed{\Delta P_{\mathrm{fuel}}=8.0\text{ kW}} \]

This question demonstrates an important distinction.

Immediately after the roof box is installed, the driving power has not changed, but the resistive force has increased. The vehicle therefore slows down.

To maintain the original speed again, the engine must deliver greater wheel power. Because of energy losses, the increase in fuel input power is greater than the increase in wheel power.

Question 5

This final problem explores a subtle distinction between the speed used to calculate aerodynamic drag and the speed used to calculate mechanical power at the wheels.

Two identical vehicles travel east along a level road.

Ignore rolling resistance and drivetrain losses.

The aerodynamic drag force on a vehicle travelling at 20 m/s relative to the surrounding air is 400 N.

Consider two situations:

Vehicle A

• Travels east at 20 m/s relative to the road.
• The surrounding air is stationary.

Vehicle B

• Travels east at 30 m/s relative to the road.
• The surrounding air moves east at 10 m/s.

Assume the air density and vehicle orientation are identical.

(a) Calculate the speed of each vehicle relative to the surrounding air.

(b) Determine the aerodynamic drag force acting on each vehicle.

(c) Calculate the wheel power required to maintain constant speed in each situation.

(d) Calculate the work supplied by the wheels against aerodynamic drag when each vehicle travels 100 m along the road. How long does each vehicle take to cover this distance?

(e) Both vehicles experience the same drag force. Explain why they do not require the same wheel power.

A student argues that the power must be identical because the relative air speeds and drag forces are identical. Another student argues that Vehicle B requires 1.5 times as much power because it travels 1.5 times further each second.

Determine which student is correct and justify your conclusion.

Solution 5

The second student is correct. Vehicle B requires 12 kW, while Vehicle A requires 8 kW, even though both experience the same aerodynamic drag force of 400 N.

The key is that relative air speed determines the drag force, but road speed determines the rate at which the wheels do work against that force in the road reference frame.

Part (a)

For Vehicle A, the surrounding air is stationary.

Therefore:

\[ v_{\mathrm{air,A}}=20\text{ m/s} \]

For Vehicle B, the vehicle and air move in the same direction.

The relative air speed is:

\[ v_{\mathrm{air,B}}=30-10 \]

\[ v_{\mathrm{air,B}}=20\text{ m/s} \]

Therefore, both vehicles have the same speed relative to the surrounding air.

Part (b)

The drag force is determined by the relative air speed, assuming the same vehicle shape and air density.

Both vehicles experience:

\[ \boxed{F_D=400\text{ N}} \]

Part (c)

This is where the two velocities must be distinguished.

The mechanical power delivered at the wheels is:

\[ P_{\mathrm{wheel}}=F_Dv_{\mathrm{road}} \]

For Vehicle A:

\[ P_A=400\times20 \]

\[ \boxed{P_A=8000\text{ W}=8\text{ kW}} \]

For Vehicle B:

\[ P_B=400\times30 \]

\[ \boxed{P_B=12000\text{ W}=12\text{ kW}} \]

Therefore:

\[ \frac{P_B}{P_A}=\frac{12}{8} \]

\[ \boxed{\frac{P_B}{P_A}=1.5} \]

Vehicle B requires 1.5 times the wheel power.

Part (d)

Now consider the work done over the same road distance of 100 m.

Work is:

\[ W=Fd \]

For Vehicle A:

\[ W_A=400\times100 \]

\[ \boxed{W_A=40000\text{ J}=40\text{ kJ}} \]

For Vehicle B:

\[ W_B=400\times100 \]

\[ \boxed{W_B=40\text{ kJ}} \]

Both vehicles require the same amount of work against drag over the same road distance.

However, they take different amounts of time.

For Vehicle A:

\[ t_A=\frac{d}{v} \]

\[ t_A=\frac{100}{20} \]

\[ \boxed{t_A=5.0\text{ s}} \]

For Vehicle B:

\[ t_B=\frac{100}{30} \]

\[ \boxed{t_B=3.33\text{ s}} \]

Vehicle B completes the same amount of work in less time.

This explains its greater power requirement.

We can verify the results using \(P=W/t\).

For Vehicle A:

\[ P_A=\frac{40000}{5.0} \]

\[ P_A=8.0\text{ kW} \]

For Vehicle B:

\[ P_B=\frac{40000}{3.33} \]

\[ P_B\approx12.0\text{ kW} \]

The results agree with our earlier calculations.

Part (e)

The second student is correct because power depends on both force and the speed of the object on which that force acts.

The tempting mistake is to use relative air speed in every equation simply because aerodynamic drag depends on relative air speed.

However, these equations describe different things:

• The relative air speed determines the magnitude of aerodynamic drag.
• The vehicle’s road speed determines how quickly the vehicle moves through a road distance while that drag force opposes it.

For the wheel-power calculation, we have consistently used the road as our reference frame.

Over the same distance, both vehicles require 40 kJ of work against drag. Over the same time, however, Vehicle B covers more distance and therefore requires more energy.

For example, over six seconds:

Vehicle A travels:

\[ d_A=20\times6=120\text{ m} \]

Its work against drag is:

\[ W_A=400\times120=48\text{ kJ} \]

Vehicle B travels:

\[ d_B=30\times6=180\text{ m} \]

Its work against drag is:

\[ W_B=400\times180=72\text{ kJ} \]

Vehicle B transfers 1.5 times as much energy over the same six-second interval.

The apparent contradiction disappears once we separate the amount of work done from the rate at which it is done.

This is the deeper reason that greater power does not necessarily mean greater force, and equal force does not necessarily mean equal power.

When solving vehicle-power problems, always identify the force acting on the vehicle, the vehicle’s speed in the chosen reference frame, and whether the question asks for energy transferred or power.

That distinction is what connects work, resistive forces, constant-speed motion, and power into one consistent model.