How to Calculate Overtaking Time Using Relative Velocity

Two vehicles are travelling east along a straight road. A hatchback is 90 m behind a van, but the hatchback is travelling at \(24\,\mathrm{m\,s^{-1}}\) while the van travels at \(18\,\mathrm{m\,s^{-1}}\).

Will the hatchback catch the van in \(90 \div 24 = 3.75\) seconds?

No. The van keeps moving. Every second, the hatchback travels 24 m while the van travels another 18 m. The gap shrinks by only 6 m.

To find the overtaking time, divide the initial gap by the rate at which the faster vehicle gains on the slower one:

\[ t=\frac{90}{24-18}=15\,\mathrm{s} \]

The hatchback draws level with the van after 15 seconds.

That’s the basic method for two vehicles moving at constant velocities in the same direction: initial gap divided by relative closing speed. The difficult part is recognising when that shortcut works, when it doesn’t, and exactly what counts as overtaking.

01Choose the observer before subtracting velocities

We need two reference frames.

• Ground frame (G): An observer standing beside the road measures both vehicles’ positions and velocities relative to the stationary road.
• Van frame (B): An observer travelling with the van measures the hatchback’s position and velocity relative to the van.

The hatchback is object A. We’ll define east as positive and west as negative.

From the ground observer’s perspective:

\[ \begin{aligned} v_{A/G}&=+24\,\mathrm{m\,s^{-1}}\\ v_{B/G}&=+18\,\mathrm{m\,s^{-1}} \end{aligned} \]

The symbol \(v_{A/G}\) means the signed east-west component of A’s velocity measured in the ground frame. Similarly, \(v_{B/G}\) is B’s velocity relative to the ground.

Now imagine sitting in the van. The van appears stationary, while the hatchback approaches from behind.

Its velocity relative to the van is:

\[ v_{A/B}=v_{A/G}-v_{B/G} \]

Substituting:

\[ \begin{aligned} v_{A/B}&=24-18\\ &=+6\,\mathrm{m\,s^{-1}} \end{aligned} \]

The positive sign tells us the hatchback moves east relative to the van. Because it starts behind the van, that movement reduces their separation.

Relative velocity is a vector, not just a speed difference

Velocity has both magnitude and direction. Speed has magnitude only.

In this one-dimensional example, we represented each velocity vector by a signed number. The hatchback’s relative velocity component is \(+6\,\mathrm{m\,s^{-1}}\), and the magnitude of that velocity, its relative speed, is \(6\,\mathrm{m\,s^{-1}}\).

If the hatchback instead travelled at \(12\,\mathrm{m\,s^{-1}}\) east, its relative velocity would be:

\[ v_{A/B}=12-18=-6\,\mathrm{m\,s^{-1}} \]

Its relative speed would still be \(6\,\mathrm{m\,s^{-1}}\), but the negative velocity means it moves west relative to the van. Since it starts behind, the gap increases.

So a relative speed of 6 m/s does not automatically mean a closing speed of 6 m/s. You need the sign and the vehicles’ initial positions.

This is why subtracting velocity vectors is the reliable method. Subtracting scalar speeds only happens to give the correct closing rate for same-direction motion when the rear vehicle is faster.

02The overtaking-time equation

Let A be the following vehicle and B the leading vehicle.

Suppose their initial positions are \(x_{A0}\) and \(x_{B0}\), measured along the road in metres from the same origin. The initial gap between their chosen reference points is:

\[ g_0=x_{B0}-x_{A0} \]

We’re assuming \(g_0>0\), meaning B starts ahead.

If both vehicles maintain constant velocities, their positions after time \(t\), in seconds, are:

\[ \begin{aligned} x_A(t)&=x_{A0}+v_{A/G}t\\ x_B(t)&=x_{B0}+v_{B/G}t \end{aligned} \]

They draw level when \(x_A(t)=x_B(t)\).

Rearranging gives:

\[ t_{\mathrm{level}}=\frac{g_0}{v_{A/B}} \]

Here:

• \(t_{\mathrm{level}}\) is the time until the selected reference points align, in seconds.
• \(g_0\) is their initial longitudinal separation, in metres.
• \(v_{A/B}\) is A’s signed velocity relative to B, in metres per second.

For this particular situation, the formula gives a future overtaking time only when \(v_{A/B}>0\).

If \(v_{A/B}=0\), the vehicles maintain their initial gap. If \(v_{A/B}<0\), the gap increases, provided their velocities remain unchanged.

A position-time graph makes the result visible

Return to the opening example. Choose the hatchback’s starting position as \(x=0\).

Then:

\[ \begin{aligned} x_A(t)&=24t\\ x_B(t)&=90+18t \end{aligned} \]

Both position-time graphs are straight lines.

The van’s graph starts at 90 m because it has a head start. The hatchback’s graph is steeper because the hatchback has the greater velocity.

The lines intersect when \(t=15\,\mathrm{s}\). At that instant:

\[ \begin{aligned} x_A&=24(15)=360\,\mathrm{m}\\ x_B&=90+18(15)=360\,\mathrm{m} \end{aligned} \]

Both front reference points are 360 m from the hatchback’s starting position.

Notice something useful: the hatchback travels 360 m before drawing level, not 90 m. The 90 m is the amount of distance it must gain on the van.

That’s why dividing the starting gap by the hatchback’s ground speed gives the wrong result.

03How could you measure overtaking time experimentally?

Imagine testing the calculation with two small motorised cars on parallel tracks.

The question is whether measuring the cars’ speeds at the start is enough to predict when they’ll draw level.

First design: measure the speeds

Two straight, parallel lanes are marked with a shared distance scale. Each model car has a clearly identifiable front marker.

At the chosen starting instant:

• Following car A has its front marker at \(x=0\).
• Leading car B has its front marker at \(x=1.20\,\mathrm{m}\).
• A is moving east at approximately \(0.80\,\mathrm{m\,s^{-1}}\).
• B is moving east at approximately \(0.50\,\mathrm{m\,s^{-1}}\).

A light gate spanning each lane can estimate speed by timing how long a marked section of a vehicle takes to pass through the beam. A common clock is needed to relate the two vehicles’ movements.

Assuming both speeds remain constant, how long should the experiment take?

The predicted closing speed is:

\[ v_{A/B}=0.80-0.50 =0.30\,\mathrm{m\,s^{-1}} \]

Therefore:

\[ t=\frac{1.20}{0.30} =4.0\,\mathrm{s} \]

What would the measurements show?

If both cars maintain those speeds, their front-marker positions should follow these idealised values.

These are illustrative model predictions, not measurements from a real experiment.

The important pattern is that the gap decreases by 0.30 m during each second. At 4.0 s, the two front markers should have the same eastward position.

The limitation: two speed readings aren’t enough

The first design has a weakness. A light gate gives a speed estimate near its location. It doesn’t prove that a model car maintains that speed for the entire journey.

Suppose one motor briefly slows, or the cars are still accelerating when the gates record their speeds. The predicted overtaking time could be wrong.

A better design uses a fixed overhead camera recording both front markers with the same timestamp throughout the run.

You could then:

1. Measure both cars’ positions at matching times.
2. Plot position against time for each car.
3. Compare the gradients and check whether they remain constant.
4. Identify the timestamp at which their front-marker positions become equal.

If the camera data followed the table, they would support the constant-velocity model over that interval. If the position graphs curved, you would need a model that accounts for changing velocity.

This is a general lesson about motion experiments: an instantaneous speed measurement and an average speed measurement are not interchangeable, and neither proves constant velocity by itself.

04Worked examples

05Three situations that look similar but behave differently

A useful test is to write the signed velocities before doing any subtraction.

Assume A initially lies west of B along the same straight road.

In the third situation, the relative velocity is:

\[ v_{A/B}=22-(-17) =+39\,\mathrm{m\,s^{-1}} \]

Because the vehicles are approaching from opposite directions, their closing speed equals the sum of their speed magnitudes.

They may meet, but that isn’t an overtaking event in which one vehicle passes another travelling in the same direction.

For more on the opposite-direction case, see How to Calculate Closing Speed Between Two Approaching Cars.

06Drawing level isn’t always a completed overtake

So far, we’ve treated vehicles as moving reference points. That’s useful for learning the method, but real vehicles have lengths.

Suppose a 4 m car follows a 12 m bus, and the front of the car is initially 40 m behind the rear of the bus.

Assume both travel east at constant velocities in parallel lanes, with the car faster.

There are three different moments:

• First overlap: The car’s front reaches the bus’s rear.
• Front-to-front alignment: The car’s front reaches the bus’s front.
• Full clearance: The car’s rear reaches and then moves ahead of the bus’s front.

Let \(h\) represent the initial empty gap from the following vehicle’s front to the leading vehicle’s rear. Let \(L_A\) and \(L_B\) be the following and leading vehicle lengths, respectively.

All three distances are measured along the road in metres.

For constant velocities and a positive relative closing velocity \(v_{A/B}\):

The time until first overlap is:

\[ t_{\mathrm{overlap}}=\frac{h}{v_{A/B}} \]

The time until full clearance is:

\[ t_{\mathrm{clear}} =\frac{h+L_A+L_B}{v_{A/B}} \]

Why add both lengths?

The car must first close the empty gap, then gain enough distance to move its entire length past the bus’s entire length.

The instant of equality in the second equation is the boundary of full clearance. The car is completely ahead immediately afterwards.

If the question gives a front-to-front separation instead, use the corresponding front-marker positions. Don’t mix that measurement with a front-to-rear gap.

This distinction is especially important when the question says completely overtakes, passes, or clears the vehicle ahead.

07Questions and solutions

These five questions progress from a straightforward constant-velocity calculation to more difficult problems involving changing speeds, vehicle lengths, moving observers, and evidence that initially appears to rule out an overtake.

Unless otherwise stated, use east as positive, measure all positions relative to the road, assume straight parallel lanes, and ignore vehicle lengths when treating front markers as points.

Question 1

A car A travels east at \(24\,\mathrm{m\,s^{-1}}\). A bus B, initially 80 m ahead of the car, travels east at \(16\,\mathrm{m\,s^{-1}}\).

Both maintain constant velocities.

Calculate:

a) The time until their front markers draw level.

b) The distance travelled by the car during that time.

Solution 1

The car draws level after \(10\,\mathrm{s}\) and travels \(240\,\mathrm{m}\) during that time.

Part a

Relative velocity:

\[ \begin{aligned} v_{A/B}&=24-16\\ &=+8\,\mathrm{m\,s^{-1}} \end{aligned} \]

The gap shrinks by 8 m each second.

Therefore:

\[ \begin{aligned} t&=\frac{80}{8}\\ &=10\,\mathrm{s} \end{aligned} \]

Part b

The car’s displacement is:

\[ \begin{aligned} \Delta x_A&=v_At\\ &=24(10)\\ &=240\,\mathrm{m} \end{aligned} \]

The bus travels \(16(10)=160\,\mathrm{m}\). Adding its original 80 m lead gives the same final position of 240 m.

The common mistake is dividing 80 m by 24 m/s. That calculation ignores the distance the bus travels while the car catches up.

Question 2

Two cars travel east on a straight test track.

At \(t=0\), car A is at \(x=0\), travelling steadily at \(22\,\mathrm{m\,s^{-1}}\). Car B is 120 m ahead, initially travelling at \(25\,\mathrm{m\,s^{-1}}\).

Car B maintains its speed for 8.0 s. It then brakes uniformly at \(3.0\,\mathrm{m\,s^{-2}}\) for 3.0 s, reaching \(16\,\mathrm{m\,s^{-1}}\), and maintains that speed afterwards.

Car A continues at \(22\,\mathrm{m\,s^{-1}}\) throughout.

Calculate:

a) The largest front-to-front gap between the cars, and when it occurs.

b) The time from \(t=0\) until A overtakes B.

Solution 2

The maximum gap is \(145.5\,\mathrm{m}\) at \(t=9.0\,\mathrm{s}\). Car A overtakes B at \(t=34.25\,\mathrm{s}\).

The key is to treat the motion in three stages. Using the final closing speed from the beginning would ignore the period when B is initially pulling further ahead.

Part a: Find the maximum gap

For the first 8 s, B is 3 m/s faster than A.

The gap therefore increases from 120 m to:

\[ \begin{aligned} g(8)&=120+(25-22)(8)\\ &=144\,\mathrm{m} \end{aligned} \]

During braking, B’s velocity decreases from 25 to 16 m/s.

The gap reaches its maximum when B and A have equal velocities. Before that instant, B is faster and the gap grows. Afterwards, A is faster and the gap shrinks.

Let \(\tau\) be time in seconds since B began braking.

\[ 25-3\tau=22 \]

Therefore:

\[ \tau=1.0\,\mathrm{s} \]

This occurs at \(t=9.0\,\mathrm{s}\) from the original start.

During that first second of braking, B travels:

\[ \begin{aligned} \Delta x_B&=25(1)-1.5(1)^2\\ &=23.5\,\mathrm{m} \end{aligned} \]

A travels \(22(1)=22\,\mathrm{m}\).

So the maximum gap is:

\[ \begin{aligned} g_{\max}&=144+23.5-22\\ &=145.5\,\mathrm{m} \end{aligned} \]

Part b: Find the overtaking time

Calculate the gap at the end of B’s braking, at \(t=11\,\mathrm{s}\).

During the 3 s braking interval, B travels:

\[ \begin{aligned} \Delta x_B&=25(3)-1.5(3)^2\\ &=61.5\,\mathrm{m} \end{aligned} \]

A travels:

\[ \Delta x_A=22(3)=66\,\mathrm{m} \]

Therefore the gap at 11 s is:

\[ \begin{aligned} g(11)&=144+61.5-66\\ &=139.5\,\mathrm{m} \end{aligned} \]

After 11 s, B maintains 16 m/s and A maintains 22 m/s.

Their closing speed is now:

\[ 22-16=6\,\mathrm{m\,s^{-1}} \]

The additional time to close the remaining gap is:

\[ \begin{aligned} \Delta t&=\frac{139.5}{6}\\ &=23.25\,\mathrm{s} \end{aligned} \]

Thus:

\[ \begin{aligned} t_{\mathrm{total}}&=11+23.25\\ &=34.25\,\mathrm{s} \end{aligned} \]

As a position check, A travels \(22(34.25)=753.5\,\mathrm{m}\). B’s final position is also 753.5 m.

The surprising part is that the gap keeps increasing briefly after B starts braking. Braking doesn’t instantly make B slower than A. It takes one second for their velocities to match, and only then does A begin to gain.

Question 3

A 4.0 m hatchback A travels east at \(27\,\mathrm{m\,s^{-1}}\), following a 12.0 m coach B travelling east at \(20\,\mathrm{m\,s^{-1}}\).

Both vehicles maintain constant velocities in parallel lanes.

At \(t=0\), the front bumper of A is 45.0 m behind the rear bumper of B.

Calculate:

a) When A’s front first reaches B’s rear.

b) When their front bumpers are level.

c) When A has completely cleared B, and the time taken between first overlap and full clearance.

Solution 3

First overlap occurs at approximately \(6.43\,\mathrm{s}\), front-to-front alignment at \(8.14\,\mathrm{s}\), and full clearance at \(8.71\,\mathrm{s}\). The interval from first overlap to full clearance is \(2.29\,\mathrm{s}\).

The different results arise because each event involves a different pair of reference points.

The car’s relative velocity is:

\[ \begin{aligned} v_{A/B}&=27-20\\ &=7\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part a: First overlap

The front of A initially lies 45.0 m behind the rear of B.

\[ \begin{aligned} t_{\mathrm{overlap}}&=\frac{45.0}{7}\\ &\approx6.43\,\mathrm{s} \end{aligned} \]

Part b: Front-to-front alignment

After reaching B’s rear, A must gain another 12.0 m to reach B’s front.

Total required relative displacement:

\[ 45.0+12.0=57.0\,\mathrm{m} \]

Therefore:

\[ \begin{aligned} t_{\mathrm{front}}&=\frac{57.0}{7}\\ &\approx8.14\,\mathrm{s} \end{aligned} \]

Part c: Full clearance

A’s rear must reach B’s front. That requires an additional 4.0 m of relative displacement after their fronts align.

The total required relative displacement is:

\[ 45.0+12.0+4.0=61.0\,\mathrm{m} \]

Thus:

\[ \begin{aligned} t_{\mathrm{clear}}&=\frac{61.0}{7}\\ &\approx8.71\,\mathrm{s} \end{aligned} \]

The time between first overlap and full clearance is:

\[ \begin{aligned} \Delta t&=\frac{12.0+4.0}{7}\\ &\approx2.29\,\mathrm{s} \end{aligned} \]

The tempting mistake is to stop at 8.14 s and call the overtake complete. At that instant, the hatchback’s rear is still 4.0 m behind the coach’s front. The hatchback must gain that final 4.0 m before clearing the coach.

Question 4

A patrol vehicle P travels east at \(18\,\mathrm{m\,s^{-1}}\).

At \(t=0\), P is at ground coordinate \(x=200\,\mathrm{m}\). A car A is alongside P, while a van B is 140 m farther east.

An observer inside P measures:

• Car A’s velocity relative to P as \(+2\,\mathrm{m\,s^{-1}}\).
• Van B’s velocity relative to P as \(-6\,\mathrm{m\,s^{-1}}\).

Car A and van B both maintain constant ground velocities.

Immediately after the measurements, the patrol vehicle brakes uniformly at \(1.0\,\mathrm{m\,s^{-2}}\) for 6.0 s, then maintains its new velocity.

Calculate:

a) The ground velocities of A and B.

b) When and where A overtakes B, using the ground coordinate system.

c) Whether the patrol vehicle’s braking changes the overtaking time. Explain what the patrol observer sees when the braking finishes.

Solution 4

Car A travels east at \(20\,\mathrm{m\,s^{-1}}\), van B travels east at \(12\,\mathrm{m\,s^{-1}}\), and A overtakes B after \(17.5\,\mathrm{s}\) at ground coordinate \(x=550\,\mathrm{m}\). The patrol vehicle’s braking does not change that event.

The important constraint is that A’s and B’s ground motions are unaffected by P’s braking. Only their velocities as measured from P change.

Part a: Ground velocities

Relative velocity is calculated by subtracting the observer’s velocity.

For A:

\[ v_{A/P}=v_{A/G}-v_{P/G} \]

Rearranging:

\[ \begin{aligned} v_{A/G}&=v_{A/P}+v_{P/G}\\ &=2+18\\ &=+20\,\mathrm{m\,s^{-1}} \end{aligned} \]

For B:

\[ \begin{aligned} v_{B/G}&=v_{B/P}+v_{P/G}\\ &=-6+18\\ &=+12\,\mathrm{m\,s^{-1}} \end{aligned} \]

Both vehicles move east relative to the road.

The negative value for B in the patrol frame doesn’t mean B is travelling west on the road. It means B moves backwards relative to a faster eastbound observer.

Part b: Overtaking time and position

A starts at \(x_A=200\,\mathrm{m}\).

B starts at:

\[ x_B=200+140=340\,\mathrm{m} \]

The relative velocity of A with respect to B is:

\[ \begin{aligned} v_{A/B}&=20-12\\ &=8\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} t&=\frac{140}{8}\\ &=17.5\,\mathrm{s} \end{aligned} \]

The ground coordinate where A catches B is:

\[ \begin{aligned} x_A&=200+20(17.5)\\ &=550\,\mathrm{m} \end{aligned} \]

Checking B:

\[ \begin{aligned} x_B&=340+12(17.5)\\ &=550\,\mathrm{m} \end{aligned} \]

Part c: Does the observer’s braking change anything?

No. A and B keep their constant ground velocities, so their relative velocity remains \(8\,\mathrm{m\,s^{-1}}\).

After 6.0 s, P’s ground velocity is:

\[ \begin{aligned} v_{P/G}&=18-1.0(6.0)\\ &=12\,\mathrm{m\,s^{-1}} \end{aligned} \]

At that moment, P measures:

\[ \begin{aligned} v_{A/P}&=20-12=+8\,\mathrm{m\,s^{-1}}\\ v_{B/P}&=12-12=0\,\mathrm{m\,s^{-1}} \end{aligned} \]

The van now appears stationary to the patrol observer, even though it continues moving east at 12 m/s relative to the road.

Throughout P’s braking, the measured velocities of A and B relative to P change by the same amount. Their difference remains 8 m/s.

The tempting error is to think an accelerating observer changes when two other vehicles meet. Changing the observer changes the measured positions and velocities, but not the physical coincidence event or the elapsed time in this classical model.

Question 5

An automated camera records the positions of two cars travelling east on a straight test track.

The camera takes one image at \(t=0\) and another at \(t=10.0\,\mathrm{s}\).

In both images, the front marker of car A is 16.0 m behind the front marker of car B.

The system concludes that A never overtook B during those 10 seconds.

Additional motion data establish that:

• B travels east at a constant \(20\,\mathrm{m\,s^{-1}}\).
• A initially travels east at \(30\,\mathrm{m\,s^{-1}}\).
• A decelerates uniformly at \(2.0\,\mathrm{m\,s^{-2}}\) throughout the 10 s interval.
• Both cars remain on straight, parallel lanes and continue moving east.

a) Is the camera system’s conclusion correct? Determine every instant when the front markers align, and identify which car overtakes the other.

b) Find A’s greatest lead over B, if any, and when it occurs.

c) Now suppose the initial gap is an unknown positive distance \(d\), rather than 16.0 m, while both vehicles follow the same velocity histories. Determine exactly which values of \(d\) allow A to overtake B and then be overtaken by B again during the 10 s interval.

Explain what happens at the limiting value of \(d\), and why that boundary case must be treated differently.

Solution 5

The camera system is wrong. With a 16.0 m initial gap, A overtakes B at \(t=2.0\,\mathrm{s}\), and B overtakes A again at \(t=8.0\,\mathrm{s}\). A’s greatest lead is 9.0 m at \(t=5.0\,\mathrm{s}\).

More generally, two changes in order occur when \(0<d<25\,\mathrm{m}\). At exactly \(d=25\,\mathrm{m}\), the front markers briefly align without either vehicle changing its position from behind to ahead.

The conceptual trap is assuming that matching initial and final separations imply nothing happened between the two measurements.

Part a: Determine whether overtaking occurred

Choose A’s initial position as \(x_A(0)=0\), so B starts at \(x_B(0)=16\,\mathrm{m}\).

For A, use the constant-acceleration equation:

\[ x_A=ut+\frac12at^2 \]

Substituting \(u=30\,\mathrm{m\,s^{-1}}\) and \(a=-2.0\,\mathrm{m\,s^{-2}}\):

\[ x_A=30t-t^2 \]

For B:

\[ x_B=16+20t \]

The front markers align when \(x_A=x_B\):

\[ 30t-t^2=16+20t \]

Rearranging:

\[ \begin{aligned} t^2-10t+16&=0\\ (t-2)(t-8)&=0 \end{aligned} \]

Therefore:

\[ t=2.0\,\mathrm{s},\quad t=8.0\,\mathrm{s} \]

Both solutions lie within the observed 10 s interval.

Which car is passing at each instant?

A’s velocity is:

\[ v_A=30-2t \]

At 2 s:

\[ v_A=30-2(2)=26\,\mathrm{m\,s^{-1}} \]

A is faster than B’s 20 m/s and moves from behind to ahead. A overtakes B.

At 8 s:

\[ v_A=30-2(8)=14\,\mathrm{m\,s^{-1}} \]

Now B is faster. B moves from behind A to ahead of it, overtaking A.

Both vehicles have continued east throughout. The direction of their relative motion has reversed.

Part b: Find the maximum lead

Let \(r(t)=x_A-x_B\) represent A’s signed position relative to B, in metres.

Positive \(r\) means A is ahead. Negative \(r\) means A is behind.

\[ \begin{aligned} r(t)&=(30t-t^2)-(16+20t)\\ &=-16+10t-t^2 \end{aligned} \]

A’s lead is greatest when the relative velocity becomes zero.

\[ v_{A/B}=10-2t \]

Setting this to zero:

\[ 10-2t=0 \]

Therefore \(t=5.0\,\mathrm{s}\).

At that instant:

\[ \begin{aligned} r(5)&=-16+10(5)-(5)^2\\ &=9.0\,\mathrm{m} \end{aligned} \]

A leads B by 9.0 m at 5.0 s.

Notice that A and B have equal velocities at this instant, but their positions are different. Equality of velocity identifies the maximum relative displacement, not an overtaking event.

Part c: Find the full range of initial gaps

Now let B start \(d\) metres ahead of A, where \(d>0\).

The relative position becomes:

\[ r(t)=-d+10t-t^2 \]

Complete the square:

\[ r(t)=25-d-(t-5)^2 \]

This form reveals the entire motion without requiring us to solve a different quadratic for every possible starting gap.

Because \((t-5)^2\) is never negative, the maximum value of \(r(t)\) is:

\[ r_{\max}=25-d \]

This maximum occurs at \(t=5\,\mathrm{s}\).

There are three cases.

Case 1: \(0<d<25\,\mathrm{m}\)

Here, \(r_{\max}>0\).

A begins behind B, moves ahead during the interval, and finishes behind B again.

There must be two distinct crossings of \(r=0\).

Setting \(r=0\):

\[ (t-5)^2=25-d \]

Therefore the two crossing times are:

\[ t=5\pm\sqrt{25-d} \]

With \(d\) expressed numerically in metres, these times are in seconds.

Both crossing times lie strictly between 0 and 10 s because \(0<d<25\).

So A genuinely overtakes B and is then overtaken again.

Case 2: \(d=25\,\mathrm{m}\)

Now:

\[ r(t)=-(t-5)^2 \]

The relative position is zero only at \(t=5\,\mathrm{s}\), and is negative at every other time.

At that instant, the cars have equal positions and equal velocities. However, A is behind B both immediately before and immediately after.

The front markers align momentarily, but there is no change in their order. This is a tangency on the relative-position graph, not a genuine pass.

Case 3: \(d>25\,\mathrm{m}\)

Here, \(r_{\max}<0\).

Even at A’s greatest relative gain, it remains behind B. No overtaking occurs.

The entire classification is therefore:

Finally, why are the camera’s two images misleading?

Over the full 10 s, A initially gains distance because it is faster, but later loses the same amount because it has become slower.

The net change in relative position is zero. That doesn’t mean the relative position stayed constant.

Two snapshots can therefore show the same gap even when two overtakes happened between them. Continuous position records, or enough reliable information about the vehicles’ motion, are needed to establish what happened in the interval.

08What to decide before your next overtaking question

The most important decision isn’t which numbers to subtract. It’s which event you’re actually trying to calculate.

If two vehicles travel in the same direction at constant velocities, identify the rear and leading vehicles, choose a positive direction, subtract their signed velocities, and divide the appropriate initial gap by the positive closing rate.

If speeds change, write the position equations and solve for the relevant crossing. If the vehicles have lengths, check whether the question means first overlap, front-to-front alignment, or complete clearance.

And if two vehicles are temporarily travelling at the same speed, don’t assume they’re alongside each other. Equal velocities tell you how their separation is changing at that instant, not what the separation is.

The natural next step is How to Find Relative Acceleration from Two Acceleration Vectors, where the same observer-and-object reasoning explains how relative velocity changes over time.