How to Calculate Closing Speed Between Two Approaching Cars

Two cars are driving towards each other on a straight road. Car A travels east at \(72\ \mathrm{km\,h^{-1}}\), and Car B travels west at \(54\ \mathrm{km\,h^{-1}}\). Is the distance between them shrinking at \(18\), \(72\), or \(126\ \mathrm{km\,h^{-1}}\)?

The answer is \(126\ \mathrm{km\,h^{-1}}\). Both cars are moving towards each other, so both contribute to reducing the gap. This is their closing speed, also called their speed of approach.

For two cars travelling directly towards each other along the same straight line:

\[
v_{\text{closing}}=v_A+v_B
\]

Here, \(v_A\) and \(v_B\) represent the positive magnitudes of their speeds relative to the road, not signed velocities. If the cars travel in the same direction, or their directions change, you need a more careful method.

01Why do you add the speeds?

Forget the equation for a moment. Imagine watching both cars from a fixed position beside the road.

Convert their speeds to metres per second:

\[
\begin{aligned}
v_A&=\frac{72}{3.6}=20\ \mathrm{m\,s^{-1}}\\
v_B&=\frac{54}{3.6}=15\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

In one second:

• Car A travels \(20\ \mathrm{m}\) towards B.
• Car B travels \(15\ \mathrm{m}\) towards A.
• The distance separating them decreases by \(35\ \mathrm{m}\).

Neither car actually travels \(35\ \mathrm{m}\) in that second. Their combined motion removes \(35\ \mathrm{m}\) from the gap.

This is the first useful mental model: imagine the separation as the length of a measuring tape stretched between the cars. Both ends move inwards, so its length decreases through both movements.

The analogy works while the cars approach along one straight line. It becomes incomplete once they pass each other: the physical separation cannot become negative, and it starts increasing again.

02Choose the observer before calculating

There are two different questions hidden inside a closing-speed problem:

1. How fast is each car moving relative to the road?
2. How fast is Car B moving relative to someone travelling in Car A?

The first uses a stationary roadside observer. The second uses a moving observer inside Car A.

We can calculate both, but we must keep the reference frames clear.

The ground frame

Call the road frame \(G\). Choose east as positive and west as negative.

Velocity is a vector, so its sign tells us direction.

For the original cars:

Both velocities are measured in the same ground frame.

The frame of Car A

Now imagine sitting inside Car A and watching B.

From your moving viewpoint, B appears to approach faster than it does to someone standing beside the road.

To calculate B’s velocity relative to A:

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

Here:

• \(v_{B/A}\) is B’s signed velocity relative to A.
• \(v_{B/G}\) is B’s signed velocity relative to the road.
• \(v_{A/G}\) is A’s signed velocity relative to the road.

Substitute the signed velocities:

\[
\begin{aligned}
v_{B/A}&=-15-(+20)\\
&=-35\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The negative sign indicates west.

So someone travelling in A observes B moving west at \(35\ \mathrm{m\,s^{-1}}\).

The relative speed is the magnitude of this velocity:

\[
|v_{B/A}|=35\ \mathrm{m\,s^{-1}}
\]

Because the cars are approaching directly along the same line, this is also their closing speed.

Relative velocity is not the same as relative speed

Relative velocity includes direction. Relative speed does not.

The general straight-line formula is:

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

The magnitude of the relative velocity is:

\[
s_{\text{relative}}=|v_{B/A}|
\]

Closing speed asks a further question: is the actual distance between the objects decreasing?

When B is east of A and both move directly towards each other, the relative speed equals the closing speed.

If the cars have already passed each other, they still have the same relative speed, but their separation is increasing. They now have an opening speed rather than a closing speed.

For a broader look at how direction changes relative motion, see Can Two Objects Have the Same Speed but Different Velocities?.

03How could you actually measure closing speed?

Suppose you want to test the calculation rather than simply trust two speedometer readings.

The first measurement has a problem

Imagine placing two speed-measuring devices beside a straight test road.

One displays \(20\ \mathrm{m\,s^{-1}}\). The other displays \(15\ \mathrm{m\,s^{-1}}\).

Both report only speed magnitude.

Can you conclude that the cars are approaching at \(35\ \mathrm{m\,s^{-1}}\)?

Not yet. The cars could be travelling towards each other, travelling in the same direction, or moving away from each other.

The instruments haven’t recorded direction, so the measurements cannot distinguish these situations.

An improved design: measure positions over time

Consider a proposed experiment with a fixed overhead camera, a straight marked test section, and a shared time reference.

The camera looks vertically down onto two parallel lanes. Car A is initially west of B. Both cars remain on their own lanes, allowing their east-west coordinates to be compared without requiring them to collide.

The road markings provide a position scale in metres. Each time-stamped image records both car centres in the same ground reference frame.

Suppose the following positions were recorded. These are illustrative values calculated for ideal constant velocities, not observations from a real experiment.

The camera provides information that the original speed displays could not: positions, directions, and changes in separation.

Before calculating, predict what the gap should be after one second if the two ground speeds really are \(20\) and \(15\ \mathrm{m\,s^{-1}}\) in opposite directions.

It should fall from \(140\ \mathrm{m}\) to \(105\ \mathrm{m}\), exactly as in the illustrative table.

Calculate closing speed from the measurements

For an interval during which the cars remain in the same order, use:

\[
c_{\text{avg}}=\frac{d_1-d_2}{\Delta t}
\]

where:

• \(c_{\text{avg}}\) is the average closing speed in \(\mathrm{m\,s^{-1}}\),
• \(d_1\) is the initial along-road gap in metres,
• \(d_2\) is the final along-road gap in metres, and
• \(\Delta t\) is the elapsed time in seconds.

Using the first two measurements:

\[
\begin{aligned}
c_{\text{avg}}&=\frac{140-105}{1.0}\\
&=35\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The calculation agrees with adding the ground-speed magnitudes.

This agreement supports the straight-line, constant-speed model for the illustrated interval.

However, it does not prove that the cars will continue at these velocities. If either driver brakes, accelerates, or turns, the closing speed changes.

A fixed camera recording positions at one-second intervals also gives an average closing speed over each interval. It cannot necessarily determine the exact instantaneous closing speed halfway between observations.

05When do you subtract instead?

Suppose Car A travels east at \(25\ \mathrm{m\,s^{-1}}\), while Car B is ahead of it, also travelling east, at \(18\ \mathrm{m\,s^{-1}}\).

A is faster, so the gap decreases. But both cars are moving east.

How much of the gap disappears each second?

A moves \(25\ \mathrm{m}\), but B moves another \(18\ \mathrm{m}\) away. Only \(7\ \mathrm{m}\) of the separation disappears.

Therefore:

\[
\begin{aligned}
c&=25-18\\
&=7\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The sum \(25+18\) would be wrong because the cars aren’t moving towards each other.

06Three decisions that prevent most mistakes

Two further details are worth remembering.

First, always establish the observer. You cannot directly add one car’s ground speed to a speed measured from an unrelated moving camera. Convert them to a common reference frame first.

Second, the closing speed of the distance between two objects isn’t always identical to their relative speed. On separate parallel lanes, the cars can move rapidly relative to each other while their actual straight-line distance changes more slowly.

The separation is a length, so it cannot become negative. The signed coordinate difference \(x_B-x_A\), however, can become negative when one car passes the other.

07Questions and solutions

Assume ordinary non-relativistic motion throughout. Unless a question states otherwise, use east as positive, west as negative, and velocities measured relative to a stationary road observer. Treat cars as point objects when calculating when their positions coincide.

Question 1

Car A travels east at \(14\ \mathrm{m\,s^{-1}}\), while Car B travels west at \(11\ \mathrm{m\,s^{-1}}\). They are initially \(175\ \mathrm{m}\) apart on the same straight line, with B east of A.

(a) Calculate their closing speed.

(b) Determine how long it takes for their positions to coincide, assuming constant velocities.

(c) State B’s velocity relative to A, including its direction.

Solution 1

The closing speed is \(25\ \mathrm{m\,s^{-1}}\), their positions coincide after \(7.0\ \mathrm{s}\), and B’s velocity relative to A is \(25\ \mathrm{m\,s^{-1}}\) west.

For (a), both cars reduce the gap:

\[
\begin{aligned}
c&=14+11\\
&=25\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For (b), divide the initial separation by the constant closing speed:

\[
\begin{aligned}
t&=\frac{175}{25}\\
&=7.0\ \mathrm{s}
\end{aligned}
\]

For (c), apply the signed-velocity formula:

\[
\begin{aligned}
v_{B/A}&=v_{B/G}-v_{A/G}\\
&=-11-(+14)\\
&=-25\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The negative sign indicates west.

The tempting mistake is to subtract \(14-11\) because relative velocity involves subtraction. But subtracting signed velocities already accounts for their opposing directions.

Question 2

A fixed roadside sensor measures Car A travelling east at \(20\ \mathrm{m\,s^{-1}}\).

Car B is \(180\ \mathrm{m}\) east of A, but its ground velocity is unknown.

A third vehicle, Camera Car C, travels east at \(5.0\ \mathrm{m\,s^{-1}}\). Its onboard measuring system reports that B is moving west relative to C at \(19\ \mathrm{m\,s^{-1}}\).

Assume all three vehicles maintain constant velocities along the same east-west axis.

(a) Determine B’s velocity relative to the road.

(b) Calculate B’s velocity relative to A.

(c) Determine when A and B’s east-west positions coincide.

(d) A student adds \(20+19=39\ \mathrm{m\,s^{-1}}\) to obtain the closing speed. Explain precisely why this is incorrect.

Solution 2

Car B travels west at \(14\ \mathrm{m\,s^{-1}}\) relative to the road. B moves west at \(34\ \mathrm{m\,s^{-1}}\) relative to A, and their positions coincide after approximately \(5.3\ \mathrm{s}\).

For (a), C’s camera gives a relative velocity, not a ground velocity.

Using \(v_{B/C}=-19\ \mathrm{m\,s^{-1}}\):

\[
v_{B/G}=v_{B/C}+v_{C/G}
\]

Therefore:

\[
\begin{aligned}
v_{B/G}&=-19+(+5.0)\\
&=-14\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

B’s ground velocity is \(14\ \mathrm{m\,s^{-1}}\) west.

For (b), subtract A’s ground velocity:

\[
\begin{aligned}
v_{B/A}&=-14-(+20)\\
&=-34\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Thus the closing speed is \(34\ \mathrm{m\,s^{-1}}\).

For (c):

\[
\begin{aligned}
t&=\frac{180}{34}\\
&=5.29\ \mathrm{s}
\end{aligned}
\]

The positions coincide after approximately \(5.3\ \mathrm{s}\).

For (d), the student’s calculation mixes reference frames.

The \(20\ \mathrm{m\,s^{-1}}\) reading is relative to the road, while \(19\ \mathrm{m\,s^{-1}}\) is relative to moving vehicle C.

To compare them, convert both velocities into the same frame.

Alternatively, calculate A’s velocity in C’s frame:

\[
\begin{aligned}
v_{A/C}&=20-5\\
&=+15\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Now both velocities are measured relative to C:

\[
\begin{aligned}
v_{B/A}&=v_{B/C}-v_{A/C}\\
&=-19-(+15)\\
&=-34\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Both methods agree.

The subtle point is that relative motion is independent of which uniformly moving observer you use, provided every velocity is transformed consistently.

Question 3

Two cars begin \(200\ \mathrm{m}\) apart on a straight line.

At \(t=0\):

• Car A is at \(x=0\) and travels east at \(18\ \mathrm{m\,s^{-1}}\).
• Car B is at \(x=200\ \mathrm{m}\) and travels west at \(12\ \mathrm{m\,s^{-1}}\).

Car A maintains its velocity. Car B immediately begins braking with constant acceleration opposite to its direction of motion.

An overhead tracking system reports that the along-road separation is \(104\ \mathrm{m}\) after \(4.0\ \mathrm{s}\).

Car B comes to rest at that instant and remains stationary.

(a) Determine B’s acceleration during braking, including its direction.

(b) Calculate the instantaneous closing speed at \(t=4.0\ \mathrm{s}\).

(c) Find the total elapsed time until the cars’ positions coincide, assuming A continues at constant velocity.

(d) Explain why dividing \(200\ \mathrm{m}\) by the initial closing speed gives an incorrect meeting time.

Solution 3

B’s acceleration is \(+3.0\ \mathrm{m\,s^{-2}}\) east, the closing speed at four seconds is \(18\ \mathrm{m\,s^{-1}}\), and the cars’ positions coincide approximately \(9.8\ \mathrm{s}\) after the start.

For (a), first calculate A’s displacement in four seconds:

\[
\begin{aligned}
\Delta x_A&=vt\\
&=18(4.0)\\
&=72\ \mathrm{m}
\end{aligned}
\]

The gap decreases from \(200\ \mathrm{m}\) to \(104\ \mathrm{m}\), a reduction of \(96\ \mathrm{m}\).

Since A accounts for \(72\ \mathrm{m}\), B must have travelled \(24\ \mathrm{m}\) west.

Thus B’s signed displacement is \(-24\ \mathrm{m}\).

Use:

\[
\Delta x=ut+\frac12at^2
\]

Here, \(u\) is initial signed velocity, \(a\) is signed acceleration, and \(t\) is elapsed time.

For B:

\[
\begin{aligned}
-24&=(-12)(4)+\frac12a(4)^2\\
-24&=-48+8a\\
a&=+3.0\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The acceleration points east because B is braking while travelling west.

Check B’s velocity after four seconds:

\[
\begin{aligned}
v_B&=u+at\\
&=-12+3.0(4)\\
&=0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This agrees with the stated observation that B stops.

For (b), A is still travelling east at \(18\ \mathrm{m\,s^{-1}}\), while B is stationary.

Therefore:

\[
c=18-0=18\ \mathrm{m\,s^{-1}}
\]

B being stationary does not mean the closing speed is zero. A continues towards it.

For (c), the remaining separation at four seconds is \(104\ \mathrm{m}\).

With B stationary, A takes:

\[
\begin{aligned}
\Delta t&=\frac{104}{18}\\
&=5.78\ \mathrm{s}
\end{aligned}
\]

Adding the first four seconds:

\[
t=4.0+5.78=9.78\ \mathrm{s}
\]

The total time is approximately \(9.8\ \mathrm{s}\).

For (d), the initial closing speed was:

\[
c_0=18+12=30\ \mathrm{m\,s^{-1}}
\]

The tempting calculation is:

\[
t=\frac{200}{30}=6.67\ \mathrm{s}
\]

But \(30\ \mathrm{m\,s^{-1}}\) was only the initial closing speed.

B brakes, so its contribution to closing the gap decreases. Using the initial rate for the entire journey incorrectly assumes constant relative velocity.

The tracking measurements reveal exactly why that assumption fails.

Question 4

Two cars approach one another along parallel straight lanes whose centre lines are \(12\ \mathrm{m}\) apart.

At \(t=0\), Car B is \(16\ \mathrm{m}\) east of Car A when their positions are compared along the road.

Car A travels east at \(15\ \mathrm{m\,s^{-1}}\), while B travels west at \(5.0\ \mathrm{m\,s^{-1}}\). Both maintain constant velocities and remain on their respective lane centre lines.

A rangefinder mounted on A measures the actual straight-line distance between the car centres.

(a) Determine the magnitude of B’s velocity relative to A.

(b) Find the actual centre-to-centre distance at \(t=0\), then calculate the instantaneous rate at which this distance is decreasing.

(c) Determine when the cars are directly alongside one another and their minimum centre-to-centre distance.

(d) Calculate the average rate at which their centre-to-centre distance decreases between \(t=0\) and the instant they are alongside one another.

(e) Explain why the relative speed remains unchanged when the rangefinder’s instantaneous rate of change becomes zero.

Solution 4

The relative speed is \(20\ \mathrm{m\,s^{-1}}\), but the rangefinder initially measures a closing rate of only \(16\ \mathrm{m\,s^{-1}}\). The minimum distance is \(12\ \mathrm{m}\), reached after \(0.80\ \mathrm{s}\).

For (a), the signed ground velocities are:

\[
v_A=+15,\qquad v_B=-5
\]

Both are in \(\mathrm{m\,s^{-1}}\).

So:

\[
\begin{aligned}
v_{B/A}&=-5-(+15)\\
&=-20\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The relative speed is \(20\ \mathrm{m\,s^{-1}}\).

For (b), the initial centre-to-centre distance is the hypotenuse of a right triangle:

\[
\begin{aligned}
R&=\sqrt{16^2+12^2}\\
&=\sqrt{400}\\
&=20\ \mathrm{m}
\end{aligned}
\]

Here, \(R\) represents the actual straight-line distance measured by the rangefinder.

The relative velocity is horizontal, but the line connecting the cars is diagonal.

Only the component of relative velocity along that diagonal changes the range instantaneously.

The horizontal fraction of the diagonal is:

\[
\frac{16}{20}=0.80
\]

Therefore, the instantaneous range-closing rate is:

\[
\begin{aligned}
c_R&=20\left(\frac{16}{20}\right)\\
&=16\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The range initially decreases at \(16\ \mathrm{m\,s^{-1}}\), not \(20\ \mathrm{m\,s^{-1}}\).

For (c), the along-road gap closes at \(20\ \mathrm{m\,s^{-1}}\):

\[
\begin{aligned}
t&=\frac{16}{20}\\
&=0.80\ \mathrm{s}
\end{aligned}
\]

At that instant, the cars are directly alongside each other.

Their horizontal separation is zero, but their transverse separation remains \(12\ \mathrm{m}\).

Thus:

\[
R_{\min}=12\ \mathrm{m}
\]

For (d), the actual range decreases from \(20\ \mathrm{m}\) to \(12\ \mathrm{m}\) over \(0.80\ \mathrm{s}\).

The average range-closing rate is:

\[
\begin{aligned}
c_{\text{avg}}&=\frac{20-12}{0.80}\\
&=10\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This is smaller than the initial instantaneous rate of \(16\ \mathrm{m\,s^{-1}}\), because the angle between the line of sight and the relative velocity changes continuously.

For (e), when the cars are directly alongside one another, the line joining their centres points straight across the lanes.

Their relative velocity remains horizontal.

These directions are perpendicular, so the relative velocity has no component along the line of sight at that instant.

The rangefinder therefore records an instantaneous rate of change of zero, even though the relative speed remains \(20\ \mathrm{m\,s^{-1}}\).

Immediately afterwards, their distance starts increasing.

The tempting assumption is that relative speed always equals the rate of change of distance. It does for direct one-dimensional approach, but not for every spatial arrangement.

Question 5

Two cars move towards one another along the same idealised straight track.

Initially:

• Car A is at \(x=0\) and travels east at \(30\ \mathrm{m\,s^{-1}}\).
• Car B is at \(x=90\ \mathrm{m}\) and travels west at \(10\ \mathrm{m\,s^{-1}}\).

At \(t=0\), both drivers begin braking.

Each car experiences a constant braking acceleration of magnitude \(5.0\ \mathrm{m\,s^{-2}}\), directed opposite to its velocity while moving. Once a car stops, it remains stationary rather than reversing.

Ignore car dimensions and any interaction between the cars until their centres coincide.

A student argues:

“Initially the closing speed is \(40\ \mathrm{m\,s^{-1}}\). Both cars decelerate at \(5.0\ \mathrm{m\,s^{-2}}\), so the closing speed decreases by \(10\ \mathrm{m\,s^{-1}}\) every second. After four seconds it reaches zero. Therefore, the cars cannot meet.”

(a) Determine when each car stops if its motion is not interrupted.

(b) Test the student’s claim. Do the positions of the cars coincide? If so, at what time?

(c) Determine the closing speed at \(t=3.0\ \mathrm{s}\) and at the instant their positions first coincide.

(d) Find the critical initial separation at which the cars would just come to rest at the same position. State what initial separation is required to prevent their positions ever coinciding.

(e) Explain exactly why the student’s argument fails, even though the initial closing speed and both braking accelerations were correctly identified.

Solution 5

The cars’ positions coincide at \(t=4.0\ \mathrm{s}\), while A is still moving east at \(10\ \mathrm{m\,s^{-1}}\). The student’s claim is false because B stops after only two seconds, so its braking acceleration cannot continue changing its velocity for the full four seconds.

For (a), calculate each stopping time using:

\[
v=u+at
\]

For A, \(u=+30\ \mathrm{m\,s^{-1}}\) and \(a=-5.0\ \mathrm{m\,s^{-2}}\).

At rest:

\[
\begin{aligned}
0&=30-5.0t_A\\
t_A&=6.0\ \mathrm{s}
\end{aligned}
\]

For B, \(u=-10\ \mathrm{m\,s^{-1}}\) and \(a=+5.0\ \mathrm{m\,s^{-2}}\).

\[
\begin{aligned}
0&=-10+5.0t_B\\
t_B&=2.0\ \mathrm{s}
\end{aligned}
\]

A would stop after six seconds, but B stops after only two.

That difference is crucial.

For (b), first examine the motion during the initial two seconds, while both cars are braking and moving towards each other.

A’s displacement is:

\[
\begin{aligned}
\Delta x_A&=30(2)-\frac12(5)(2)^2\\
&=60-10\\
&=50\ \mathrm{m}
\end{aligned}
\]

B’s displacement is:

\[
\begin{aligned}
\Delta x_B&=-10(2)+\frac12(5)(2)^2\\
&=-20+10\\
&=-10\ \mathrm{m}
\end{aligned}
\]

So at \(t=2.0\ \mathrm{s}\):

\[
x_A=50\ \mathrm{m}
\]

\[
x_B=90-10=80\ \mathrm{m}
\]

Their remaining separation is:

\[
80-50=30\ \mathrm{m}
\]

B is now stationary at \(x=80\ \mathrm{m}\).

A is still moving east. Its speed at two seconds is:

\[
\begin{aligned}
v_A&=30-5(2)\\
&=20\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Now consider the following two seconds.

A’s additional eastward displacement is:

\[
\begin{aligned}
\Delta x_A&=20(2)-\frac12(5)(2)^2\\
&=40-10\\
&=30\ \mathrm{m}
\end{aligned}
\]

That is exactly the remaining gap.

Therefore, A reaches B’s stationary position at:

\[
t=2.0+2.0=4.0\ \mathrm{s}
\]

The cars’ centres first coincide after four seconds.

For (c), at \(t=3.0\ \mathrm{s}\), A’s velocity is:

\[
\begin{aligned}
v_A&=30-5(3)\\
&=15\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

B has already stopped.

Thus:

\[
c=15-0=15\ \mathrm{m\,s^{-1}}
\]

At the instant their positions coincide, \(t=4.0\ \mathrm{s}\):

\[
\begin{aligned}
v_A&=30-5(4)\\
&=10\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

B remains stationary, so the closing speed immediately before their positions coincide is:

\[
c=10\ \mathrm{m\,s^{-1}}
\]

Notice the reference-frame detail: B is stationary relative to the road, but an observer inside A still sees B approaching westwards at \(10\ \mathrm{m\,s^{-1}}\).

For (d), calculate how far each car travels before stopping, if allowed to complete its braking motion.

The stopping-distance equation is:

\[
v^2=u^2+2as
\]

where \(s\) is signed displacement.

For A:

\[
\begin{aligned}
0&=30^2+2(-5)s_A\\
s_A&=90\ \mathrm{m}
\end{aligned}
\]

For B, working with the magnitude of its westward stopping distance:

\[
\begin{aligned}
0&=10^2-2(5)d_B\\
d_B&=10\ \mathrm{m}
\end{aligned}
\]

A would travel \(90\ \mathrm{m}\) east, while B would travel \(10\ \mathrm{m}\) west.

The critical initial separation is therefore:

\[
d_{\text{critical}}=90+10=100\ \mathrm{m}
\]

At exactly \(100\ \mathrm{m}\) of initial separation, both cars would come to rest at the same position. They would just meet at rest.

For initial separations greater than \(100\ \mathrm{m}\), their stopping positions remain separated.

For smaller separations, their positions coincide before both cars have stopped.

Thus, to prevent the idealised car centres from ever coinciding, the initial separation must be greater than \(100\ \mathrm{m}\).

For (e), the student’s formula is initially correct:

\[
c=40-10t
\]

But it applies only while both cars are moving towards one another and braking at the stated rates.

At \(t=2.0\ \mathrm{s}\), B reaches zero velocity.

If its acceleration of \(+5.0\ \mathrm{m\,s^{-2}}\) continued beyond that instant, B would begin moving east. But the question explicitly says that B remains stopped.

From \(t=2.0\ \mathrm{s}\) onwards, B’s ground velocity is zero. The closing speed is simply A’s remaining eastward speed, which falls by \(5.0\ \mathrm{m\,s^{-1}}\) each second.

The corrected piecewise closing-speed model, valid before the centres coincide, is:

\[
c=
\begin{cases}
40-10t,&0\leq t\leq2,\\
30-5t,&2<t<4.
\end{cases}
\]

The second expression follows from A’s velocity \(v_A=30-5t\) and B’s stationary velocity.

At \(t=4.0\ \mathrm{s}\), the correct closing speed immediately before meeting is \(10\ \mathrm{m\,s^{-1}}\), not zero.

The student’s mistake was not arithmetic. It was applying a valid relationship beyond the physical condition under which it was valid.

This is the key lesson behind harder closing-speed problems: knowing the initial velocities is not enough when the motion changes. Track which objects are still moving, which reference frame the velocities belong to, and whether the separation is actually decreasing.

The next useful distinction is between motion along a straight line and motion along a curved path. On a bend, velocity changes direction even if speed remains constant. That leads to a different analysis of acceleration and forces, developed in Cars on Flat Circular Bends: Friction and Centripetal Force.