How to Solve Braking-Distance Problems with Reaction Time

A car is travelling towards a barrier 40 m away. Once its brakes are applied, it needs only 30 m to stop. Will it stop before the barrier?

You might predict yes. There’s 40 m available, and braking takes 30 m.

But suppose the driver takes 1.0 s to react while travelling at 15 m/s. The car covers another 15 m before braking even begins. Its total stopping distance is therefore 15 + 30 = 45 m. Without the barrier, it would stop 5 m beyond it.

The solution is to split the motion into two stages: reaction distance, when the vehicle continues at constant speed, and braking distance, when it slows with uniform negative acceleration. Calculate each distance separately, then add them.

01The two-stage method for stopping-distance problems

Picture what happens when a driver spots an obstacle.

The driver sees the obstacle, decides to brake, and moves their foot onto the brake pedal. During this interval, the car is still moving forwards.

Only after the brakes take effect does the car begin slowing down.

These are two different types of motion, so they need different equations.

Here:

• \(u\) is the initial speed, measured in metres per second (m/s).
• \(t_r\) is the reaction time, measured in seconds (s).
• \(a\) is the signed acceleration, measured in metres per second squared (m/s²).
• \(b\) is the positive magnitude of the braking deceleration, also measured in m/s².
• \(d_r\) is the reaction distance, measured in metres (m).
• \(d_b\) is the braking distance, measured in metres (m).

Adding the two distances gives:

\[ d_{\mathrm{stop}}=d_r+d_b \]

Therefore, for constant speed during reaction and uniform braking:

\[ d_{\mathrm{stop}}=ut_r+\frac{u^2}{2b} \]

This is the main result you’ll use, but knowing where it comes from is more useful than memorising it.

Stage 1: Calculate the reaction distance

During the reaction stage, the brakes haven’t taken effect.

We assume the car continues at its original speed, so its acceleration is zero.

For constant velocity:

\[ d_r=ut_r \]

Suppose a car travels at 20 m/s and the driver reacts after 0.8 s.

The car travels:

\[ \begin{aligned} d_r&=(20\,\mathrm{m/s})(0.8\,\mathrm{s})\\ &=16\,\mathrm{m} \end{aligned} \]

The driver has used 16 m of road without slowing down.

This is sometimes called thinking distance. For these problems, reaction distance and thinking distance describe the same modelled stage.

Stage 2: Calculate the braking distance

Once braking begins, we assume the car slows at a constant rate until its velocity reaches zero.

Use the uniform-acceleration equation:

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

Here, \(v\) is final velocity in m/s, and \(s\) is displacement in metres.

Choose the car’s original direction of travel as positive.

During braking:

• Initial velocity \(u\) is positive.
• Final velocity \(v=0\).
• Acceleration \(a\) is negative.
• Displacement \(s\) is positive because the car still moves forwards.

At the instant the car stops, substitute \(v=0\):

\[ 0=u^2+2as \]

Rearranging:

\[ s=-\frac{u^2}{2a} \]

Since \(a=-b\), where \(b\) is positive:

\[ d_b=\frac{u^2}{2b} \]

The braking distance is positive, even though acceleration is negative.

That distinction is important. A car moving forwards with negative acceleration is slowing down, not necessarily moving backwards.

For a refresher on why displacement can be signed while distance is always non-negative, see Distance and Displacement in HSC Physics Explained Clearly.

Why the speed-time graph has two shapes

Before braking, the speed-time graph is horizontal. The area beneath that section is a rectangle:

\[ d_r=ut_r \]

During braking, speed decreases uniformly from \(u\) to zero. The area beneath this section is a triangle.

The braking time, \(t_b\), follows from:

\[ v=u+at \]

Using \(v=0\) and \(a=-b\):

\[ t_b=\frac{u}{b} \]

The triangular area is:

\[ \begin{aligned} d_b&=\frac12ut_b\\ &=\frac12u\left(\frac{u}{b}\right)\\ &=\frac{u^2}{2b} \end{aligned} \]

So the graph and the equations describe exactly the same two-stage model.

One detail worth remembering: both stages begin with the same speed \(u\). The car hasn’t slowed during the reaction interval.

02What if the question gives the obstacle distance instead?

Sometimes the question doesn’t ask how far the car will travel. It asks what acceleration is needed to avoid an obstacle.

The obstacle distance contains both stages, so you cannot use the entire available distance as the braking displacement.

First subtract the reaction distance.

If an obstacle is \(D\) metres ahead, the distance available for braking is:

\[ d_{\mathrm{available}}=D-ut_r \]

Provided the car hasn’t already reached the obstacle before braking, the minimum braking magnitude needed to stop at the obstacle is:

\[ b_{\min}=\frac{u^2}{2d_{\mathrm{available}}} \]

Here \(D\) is the obstacle’s distance from the initial observation point, and \(b_{\min}\) is the minimum positive magnitude of uniform deceleration required.

A braking magnitude greater than this gives a shorter braking distance. A smaller magnitude gives a longer one.

03How could we test the two-stage model?

The calculations make a specific prediction: a vehicle should travel at roughly constant speed before its brakes engage, then lose equal amounts of speed in equal time intervals if its braking acceleration is uniform.

How would we tell whether that actually happens?

A model-trolley experiment

Consider a proposed low-speed experiment using a motorised trolley on a straight track.

An overhead camera faces a graduated measuring strip running alongside the trolley’s path. A warning lamp is positioned near the beginning of the measured section, and an operator has a button connected to the trolley’s braking controller.

The trolley moves from left to right.

As it passes the zero-position marker, the warning lamp illuminates. The operator responds by pressing the button. The controller activates the trolley’s brake, and the camera records the trolley’s position as it slows to rest.

The sequence is:

1. The warning lamp flashes as the trolley passes the starting marker.
2. The trolley continues moving while the operator responds.
3. The operator presses the brake button.
4. The brake engages, and the trolley slows.
5. The trolley reaches rest.

The measurement problem in the first design

Suppose the apparatus records only the warning flash and the button press.

That would measure the operator’s response interval, but not necessarily the entire period before the trolley begins slowing.

The brake controller or mechanical actuator might take additional time to respond.

To distinguish these effects, the design needs a brake-engagement indicator that changes state when the brake actually begins acting.

The camera can then identify three distinct events: the warning, the operator’s response, and brake engagement.

For the stopping-distance calculation, the relevant delay is the whole interval during which the trolley continues at its original speed. That may include both the operator’s reaction and an actuator delay.

Predicting the measurements

Suppose the trolley moves initially at 2.0 m/s, continues at that speed for 0.50 s after the warning, and then slows uniformly at \(1.0\,\mathrm{m/s^2}\).

The following values are illustrative predictions from the model, not measurements from a performed experiment.

We can verify the predicted distances.

During the reaction and actuator-delay interval:

\[ d_r=2.0(0.50)=1.0\,\mathrm{m} \]

During braking:

\[ \begin{aligned} d_b&=\frac{2.0^2}{2(1.0)}\\ &=2.0\,\mathrm{m} \end{aligned} \]

The predicted total is 3.0 m.

Notice the braking motion: the trolley travels 1.5 m during the first second of braking but only 0.5 m during the next second. Its speed is falling.

If an actual recording showed approximately equal position changes in equal time intervals before brake engagement, followed by a roughly linear decrease in speed, it would support the two-stage approximation.

It wouldn’t establish that acceleration was perfectly uniform. More video frames would be needed to test that prediction properly.

If the measured braking acceleration changed substantially, the model would need revision. The motion could be divided into shorter intervals over which acceleration is approximately uniform.

This also shows why real stopping distances cannot be predicted perfectly from a single constant deceleration. Brake response, road grip, and changing conditions can all affect the motion.

04Getting signs, roots, and physical meaning right

The most dangerous mistake in a stopping-distance problem isn’t always arithmetic. Sometimes the algebra is correct, but the answer describes a time or position the vehicle never physically reaches.

A negative acceleration does not mean negative distance

Suppose a car travels forwards at 12 m/s and brakes at \(-3.0\,\mathrm{m/s^2}\).

Its braking displacement is:

\[ \begin{aligned} 0&=12^2+2(-3.0)s\\ 0&=144-6s\\ s&=24\,\mathrm{m} \end{aligned} \]

The result is positive because the vehicle still moves forwards while slowing.

If you obtain \(-24\,\mathrm{m}\) for this situation, examine how you’ve used the acceleration sign.

An obstacle may be reached before braking starts

Suppose an obstacle is 10 m away, but the vehicle travels 14 m during the reaction interval.

The vehicle reaches the obstacle before the brakes engage.

There is no braking-stage displacement to calculate before impact.

For an obstacle \(D\) metres ahead, with \(0<D\leq d_r\), the time to reach it under the constant-speed reaction model is:

\[ t=\frac{D}{u} \]

Its speed at that point remains \(u\).

This is why checking the available distance before applying a braking formula is so useful.

Why a quadratic can give two times

If you need the time at which a braking vehicle reaches a particular point, use:

\[ s=u\tau+\frac12a\tau^2 \]

Here \(\tau\) is the time elapsed since braking began, not the time since the obstacle was first noticed.

With \(a=-b\):

\[ s=u\tau-\frac12b\tau^2 \]

Rearranging can produce a quadratic equation with two roots.

For a vehicle that brakes to rest and stays stationary, a valid braking time must satisfy:

\[ 0\leq\tau\leq\frac{u}{b} \]

The upper limit is the time required to reach zero velocity.

A later root may describe what the equation predicts if the same negative acceleration continues after the car has stopped. Under that continuation, the vehicle would start moving backwards.

That isn’t what a normal stopping-distance question means.

A reliable decision process

Before calculating, identify what the question is asking.

One particularly useful habit is to mark three positions: where the hazard is noticed, where braking begins, and where the vehicle would stop if nothing obstructed it.

If those positions are correct, most of the calculation becomes straightforward.

05Questions and solutions

The questions now build from a standard two-stage calculation to situations where changing conditions, incomplete measurements, or another moving vehicle can make the obvious method fail.

Unless otherwise stated, use straight-line motion, constant speed during reaction, and the specified uniform acceleration during each braking stage. All numerical data are illustrative.

Question 1

A car is travelling at 18 m/s when its driver notices a stationary obstacle 45 m ahead.

The driver’s reaction time is 0.75 s. The car then brakes uniformly at \(-6.0\,\mathrm{m/s^2}\).

Calculate:

• The reaction distance.
• The braking distance.
• The total stopping distance and clearance before the obstacle.
• The total time from noticing the obstacle to stopping.

Solution 1

The car stops after travelling 40.5 m, leaving 4.5 m of clearance. The total stopping time is 3.75 s.

Take forwards as positive.

First calculate reaction distance:

\[ \begin{aligned} d_r&=ut_r\\ &=18(0.75)\\ &=13.5\,\mathrm{m} \end{aligned} \]

During braking, the final velocity is zero:

\[ \begin{aligned} 0&=18^2+2(-6.0)d_b\\ 0&=324-12d_b\\ d_b&=27.0\,\mathrm{m} \end{aligned} \]

The total stopping distance is:

\[ d_{\mathrm{stop}}=13.5+27.0=40.5\,\mathrm{m} \]

Therefore, the clearance is:

\[ 45-40.5=4.5\,\mathrm{m} \]

For the braking time:

\[ \begin{aligned} 0&=18-6.0t_b\\ t_b&=3.0\,\mathrm{s} \end{aligned} \]

Add the reaction interval:

\[ t_{\mathrm{stop}}=0.75+3.0=3.75\,\mathrm{s} \]

The tempting shortcut is to give 27 m as the stopping distance. But 27 m describes only the braking stage. The extra 13.5 m travelled during reaction must be included.

Question 2

A car moves at 20 m/s towards a stationary barrier 60 m away. Once its brakes engage, its acceleration is uniformly \(-5.0\,\mathrm{m/s^2}\).

In the first situation, the driver’s reaction time is 0.80 s.

In the second situation, the reaction time increases to 1.20 s, but the original speed and braking acceleration remain unchanged.

1. Determine whether the car stops before the barrier in each situation.
2. In the second situation, calculate the time after the warning when the car reaches the barrier and its speed at that instant.
3. Explain why solving the braking displacement equation gives a second mathematical time and whether that time is physically valid.

Solution 2

With a reaction time of 0.80 s, the car stops 4 m before the barrier. With a reaction time of 1.20 s, it reaches the barrier at approximately 6.32 m/s, 3.94 s after the warning. The later quadratic root is not a valid arrival time for a vehicle that stops rather than reverses.

First situation: 0.80 s reaction time

Reaction distance:

\[ d_r=20(0.80)=16\,\mathrm{m} \]

Braking distance:

\[ \begin{aligned} d_b&=\frac{20^2}{2(5.0)}\\ &=40\,\mathrm{m} \end{aligned} \]

Total stopping distance:

\[ d_{\mathrm{stop}}=16+40=56\,\mathrm{m} \]

The car stops with 4 m remaining.

Second situation: 1.20 s reaction time

The reaction distance increases:

\[ d_r=20(1.20)=24\,\mathrm{m} \]

The braking distance remains 40 m because the initial speed and braking acceleration have not changed.

Therefore:

\[ d_{\mathrm{stop}}=24+40=64\,\mathrm{m} \]

The car cannot stop before the barrier.

It has only:

\[ 60-24=36\,\mathrm{m} \]

available for braking.

Let \(\tau\) be the time after the brakes engage when the barrier is reached.

Using \(s=u\tau+\frac12a\tau^2\):

\[ 36=20\tau-2.5\tau^2 \]

Rearrange:

\[ 2.5\tau^2-20\tau+36=0 \]

The quadratic formula gives:

\[ \tau=2.735\,\mathrm{s} \]

or:

\[ \tau=5.265\,\mathrm{s} \]

Now check the physical interval.

The car’s braking time to rest, if unobstructed, would be:

\[ t_b=\frac{20}{5.0}=4.0\,\mathrm{s} \]

The later root, 5.265 s, is beyond that stopping time. It comes from extending the uniform negative acceleration beyond zero velocity, which would make the car reverse direction.

The valid arrival time after braking begins is 2.735 s.

The total time after the warning is:

\[ \begin{aligned} t&=1.20+2.735\\ &=3.935\,\mathrm{s} \end{aligned} \]

The car’s velocity at that point is:

\[ \begin{aligned} v&=u+a\tau\\ &=20-5.0(2.735)\\ &\approx6.32\,\mathrm{m/s} \end{aligned} \]

So the car reaches the barrier at approximately 6.32 m/s.

The crucial change between the two situations is not the braking distance. It is where braking begins. An extra 0.40 s of reaction moves that starting point 8 m closer to the barrier.

Question 3

A laboratory is testing a motorised trolley with an electronically controlled brake.

A warning signal is triggered while the trolley moves at constant speed. After a delay, the brake engages and produces uniform deceleration until the trolley stops.

Two hypothetical trials give these results:

Assume the delay between warning and brake engagement is the same in both trials, and the braking deceleration magnitude is also the same.

1. Calculate the delay and the braking deceleration magnitude.
2. Predict the total stopping distance for a third trial at 24 m/s.
3. A technician argues that the delay must be zero and that the first trial can be explained entirely by braking. Show why this explanation fails when both trials are considered.
4. Explain whether these measurements alone prove that the delay is entirely a human reaction time.

Solution 3

The common delay is 1.0 s, and the braking deceleration magnitude is \(6.0\,\mathrm{m/s^2}\). The predicted stopping distance at 24 m/s is 72 m. Both trials contradict the zero-delay interpretation, and the distances alone cannot separate human reaction time from actuator delay.

Part 1: Determine the two unknowns.

Let \(r\) be the delay in seconds, and \(b\) the positive braking deceleration magnitude in m/s².

For each trial:

\[ d_{\mathrm{stop}}=ur+\frac{u^2}{2b} \]

For Trial A:

\[ 24=12r+\frac{72}{b} \]

For Trial B:

\[ 45=18r+\frac{162}{b} \]

Multiply the first equation by 1.5:

\[ 36=18r+\frac{108}{b} \]

Subtract this equation from the equation for Trial B:

\[ \begin{aligned} 45-36&=\frac{162-108}{b}\\ 9&=\frac{54}{b}\\ b&=6.0\,\mathrm{m/s^2} \end{aligned} \]

Substitute into the first equation:

\[ \begin{aligned} 24&=12r+\frac{72}{6}\\ 24&=12r+12\\ r&=1.0\,\mathrm{s} \end{aligned} \]

The signed acceleration during braking is therefore \(-6.0\,\mathrm{m/s^2}\), taking the original travel direction as positive.

Part 2: Predict the third trial.

At \(u=24\,\mathrm{m/s}\):

Reaction distance:

\[ d_r=24(1.0)=24\,\mathrm{m} \]

Braking distance:

\[ \begin{aligned} d_b&=\frac{24^2}{2(6.0)}\\ &=48\,\mathrm{m} \end{aligned} \]

Total:

\[ d_{\mathrm{stop}}=24+48=72\,\mathrm{m} \]

Part 3: Test the zero-delay explanation.

If all 24 m in Trial A were braking distance:

\[ \begin{aligned} 24&=\frac{12^2}{2b}\\ b&=3.0\,\mathrm{m/s^2} \end{aligned} \]

At 18 m/s, the same proposed deceleration would predict:

\[ \begin{aligned} d_b&=\frac{18^2}{2(3.0)}\\ &=54\,\mathrm{m} \end{aligned} \]

But Trial B has a total stopping distance of only 45 m.

The zero-delay model with one common braking acceleration cannot explain both results.

A single trial cannot determine two unknown quantities uniquely. Using different initial speeds allows the reaction and braking contributions to be separated, provided their assumed common values are valid.

Part 4: Interpret what the delay measures.

The calculated 1.0 s is the effective interval before braking begins.

It could include human response time, controller latency, or mechanical brake-engagement delay.

Without a separate measurement of the button press and actual brake engagement, calling the entire 1.0 s a human reaction time would go beyond the evidence.

Question 4

A car travels at 20 m/s along a straight road.

The driver notices a stationary barrier 85 m ahead. The road is dry from the car’s initial position to the 40 m marker, then wet from that marker onwards.

The car’s braking acceleration depends on the surface:

• Dry road: \(-5.0\,\mathrm{m/s^2}\)
• Wet road: \(-2.0\,\mathrm{m/s^2}\)

Assume speed remains constant during reaction, acceleration is uniform within each braking section, and the change in braking acceleration occurs immediately at the surface boundary.

In the first situation, the driver’s reaction time is 1.0 s.

In the second situation, the driver reacts in 0.50 s instead.

1. Determine whether the car stops before the barrier in the first situation. If not, calculate its speed at the barrier.
2. Determine the stopping position and barrier clearance in the second situation.
3. A student claims the 0.50 s improvement can save only 10 m because the car travels at 20 m/s during reaction. Explain why the change in projected stopping position is larger than 10 m.

Solution 4

With a 1.0 s reaction time, the car reaches the barrier at approximately 4.47 m/s. With a 0.50 s reaction time, it stops at the 65 m marker, leaving 20 m of clearance. The projected stopping position changes by 25 m because earlier braking also increases the distance spent slowing on the dry road.

The important constraint is that the road surface changes at a fixed position, \(x=40\,\mathrm{m}\).

Part 1: Reaction time of 1.0 s.

The reaction distance is:

\[ d_r=20(1.0)=20\,\mathrm{m} \]

Braking begins at the 20 m marker.

Only 20 m of dry road remains before the wet section starts.

Find the speed at the 40 m marker using \(v^2=u^2+2as\):

\[ \begin{aligned} v^2&=20^2+2(-5.0)(20)\\ &=400-200\\ &=200\,\mathrm{m^2/s^2} \end{aligned} \]

The car enters the wet section at speed \(\sqrt{200}\,\mathrm{m/s}\), approximately 14.14 m/s.

Its required wet-road braking distance would be:

\[ \begin{aligned} d_{\mathrm{wet}}&=\frac{200}{2(2.0)}\\ &=50\,\mathrm{m} \end{aligned} \]

Without the barrier, the projected stopping position would be:

\[ x_{\mathrm{stop}}=40+50=90\,\mathrm{m} \]

The barrier is at 85 m, so the car reaches it before stopping.

From the wet-road boundary to the barrier, 45 m remains.

\[ \begin{aligned} v^2&=200+2(-2.0)(45)\\ &=200-180\\ &=20\,\mathrm{m^2/s^2} \end{aligned} \]

Therefore:

\[ v=\sqrt{20}\approx4.47\,\mathrm{m/s} \]

The car reaches the barrier at approximately 4.47 m/s.

Part 2: Reaction time of 0.50 s.

Now the reaction distance is:

\[ d_r=20(0.50)=10\,\mathrm{m} \]

Braking begins at the 10 m marker.

The car therefore travels 30 m under the stronger dry-road braking acceleration before reaching the wet section.

The speed at 40 m satisfies:

\[ \begin{aligned} v^2&=20^2+2(-5.0)(30)\\ &=400-300\\ &=100\,\mathrm{m^2/s^2} \end{aligned} \]

So the car enters the wet road at 10 m/s.

The remaining braking distance is:

\[ \begin{aligned} d_{\mathrm{wet}}&=\frac{10^2}{2(2.0)}\\ &=25\,\mathrm{m} \end{aligned} \]

The car stops at:

\[ x_{\mathrm{stop}}=40+25=65\,\mathrm{m} \]

Its clearance before the barrier is:

\[ 85-65=20\,\mathrm{m} \]

Part 3: Explain the larger improvement.

Reacting 0.50 s earlier reduces the reaction distance by:

\[ 20(0.50)=10\,\mathrm{m} \]

But this doesn’t merely move the same braking process 10 m earlier.

It gives the car an extra 10 m of braking on the dry surface, where the deceleration magnitude is \(5.0\,\mathrm{m/s^2}\), rather than on the wet surface, where it is only \(2.0\,\mathrm{m/s^2}\).

Consequently, the car reaches the wet section at a lower speed.

The required wet braking distance drops from 50 m to 25 m.

The projected stopping position changes from 90 m to 65 m:

\[ 90-65=25\,\mathrm{m} \]

The tempting 10 m answer assumes the rest of the braking process is unchanged. That assumption fails because the surface boundary is fixed at 40 m, while the braking starting point changes.

Question 5

Two cars travel in the same direction along a straight road at 20 m/s.

The leading car’s rear bumper is initially 7.0 m ahead of the following car’s front bumper. Use these bumper positions to measure the gap between the vehicles.

At \(t=0\):

• The leading car begins braking immediately at a constant acceleration of \(-4.0\,\mathrm{m/s^2}\).
• The following driver continues at 20 m/s for 1.5 s before braking at a constant acceleration of \(-10.0\,\mathrm{m/s^2}\).

Each car stops when its velocity reaches zero and then remains stationary, unless a collision changes its motion.

Assume both vehicles follow the specified motion until their first contact.

A student calculates that each car travels exactly 50 m from its own initial position before stopping.

The student concludes:

Both cars travel the same distance, so the 7.0 m gap must remain safe throughout the event.

1. Determine whether the cars make contact.
2. If they do, calculate the time of first contact after \(t=0\) and their relative speed at that instant.
3. The position equations produce two mathematical contact times. Explain the meaning of the second root.
4. Explain why comparing only the two final stopping positions gives the wrong safety conclusion.

Solution 5

The cars do make contact, approximately 2.09 s after the leading car begins braking. At first contact, the following car is moving approximately 2.45 m/s faster than the leader. The second mathematical root describes a later crossing of the extrapolated, no-collision trajectories, not a second genuine contact.

The surprising part is that both cars would finish 7.0 m apart if their motions could continue independently without a collision.

Part 1: Check the final stopping distances.

The leading car brakes immediately at a deceleration magnitude of \(4.0\,\mathrm{m/s^2}\).

Its braking distance is:

\[ \begin{aligned} d_L&=\frac{20^2}{2(4.0)}\\ &=50\,\mathrm{m} \end{aligned} \]

The following car travels for 1.5 s before braking:

\[ d_r=20(1.5)=30\,\mathrm{m} \]

Its subsequent braking distance is:

\[ \begin{aligned} d_F&=\frac{20^2}{2(10.0)}\\ &=20\,\mathrm{m} \end{aligned} \]

So its total distance would be:

\[ 30+20=50\,\mathrm{m} \]

The student’s stopping-distance calculations are correct.

The conclusion is not.

Equal total distances do not guarantee equal positions relative to one another at every earlier instant.

Part 2: Check the gap before the follower brakes.

Choose the following car’s initial front-bumper position as \(x=0\).

The leading car’s rear bumper begins at \(x=7.0\,\mathrm{m}\).

For the first 1.5 s, the following car maintains its original speed.

Its position is:

\[ x_F=20t \]

The leader’s position while braking is:

\[ x_L=7+20t-2t^2 \]

Here \(t\) is time in seconds, and the position results are in metres.

The bumper gap is:

\[ \begin{aligned} g&=x_L-x_F\\ &=7-2t^2 \end{aligned} \]

At \(t=1.5\,\mathrm{s}\):

\[ \begin{aligned} g&=7-2(1.5)^2\\ &=2.5\,\mathrm{m} \end{aligned} \]

There has been no collision yet, but the gap has narrowed from 7.0 m to 2.5 m.

The following car now begins braking.

Part 3: Calculate the first contact time.

The leading car continues with acceleration \(-4.0\,\mathrm{m/s^2}\), so its position remains:

\[ x_L=7+20t-2t^2 \]

The following car begins braking at \(t=1.5\,\mathrm{s}\).

From its initial position, its later position can be written as:

\[ x_F=20t-5(t-1.5)^2 \]

This equation applies from the beginning of the follower’s braking stage until it would stop.

Contact occurs when the following front bumper reaches the leading rear bumper:

\[ x_F=x_L \]

Substituting:

\[ 20t-5(t-1.5)^2 =7+20t-2t^2 \]

Simplifying:

\[ 3t^2-15t+18.25=0 \]

The quadratic formula gives:

\[ t=\frac{15\pm\sqrt{6}}{6} \]

Therefore:

\[ t_1\approx2.092\,\mathrm{s} \]

and:

\[ t_2\approx2.908\,\mathrm{s} \]

Both values are inside the interval over which the independent braking equations would apply.

But the first time the bumper gap becomes zero is \(t_1\).

Therefore, first contact occurs approximately 2.09 s after the leader begins braking.

Part 4: Find the relative speed at first contact.

The leader’s velocity is:

\[ v_L=20-4t \]

At \(t=2.092\,\mathrm{s}\):

\[ \begin{aligned} v_L&=20-4(2.092)\\ &\approx11.63\,\mathrm{m/s} \end{aligned} \]

The follower’s velocity after braking begins is:

\[ v_F=20-10(t-1.5) \]

Therefore:

\[ \begin{aligned} v_F&=20-10(2.092-1.5)\\ &\approx14.08\,\mathrm{m/s} \end{aligned} \]

The following car is moving faster.

The closing speed is:

\[ \begin{aligned} v_{\mathrm{relative}}&=v_F-v_L\\ &=14.08-11.63\\ &\approx2.45\,\mathrm{m/s} \end{aligned} \]

The bumpers approach one another at approximately 2.45 m/s at first contact.

Part 5: Interpret the second root.

The quadratic predicts another position equality at approximately 2.908 s.

Why?

The following car brakes much harder than the leading car. If the vehicles could pass through one another without interacting, the follower would initially gain on the leader, then lose speed rapidly enough for the leader to pull ahead again.

The two independently calculated position curves therefore intersect twice.

But the first intersection is a physical collision.

After contact, the vehicles’ velocities and accelerations would generally change, depending on the collision. Those changes are not specified.

We cannot continue using the original independent-motion equations to predict a second real contact.

The later root belongs to a hypothetical no-collision continuation of the mathematics.

Part 6: Explain the failed stopping-position argument.

The key is that the gap changes throughout the journey.

At first, the leader slows while the follower maintains 20 m/s. The follower closes the gap.

After 1.5 s, the follower starts braking harder than the leader. However, it is still moving faster for part of that braking stage, so the gap continues shrinking.

If the independent trajectories were allowed to continue, the follower would eventually become slower and the gap would widen again.

The minimum mathematical gap occurs when their velocities become equal.

Set:

\[ 20-4t=20-10(t-1.5) \]

Solving:

\[ t=2.5\,\mathrm{s} \]

At that instant, the extrapolated separation is:

\[ \begin{aligned} g&=7-2(2.5)^2\\ &\quad+5(2.5-1.5)^2\\ &=-0.5\,\mathrm{m} \end{aligned} \]

The negative gap means the independent trajectories would overlap by 0.5 m. It is not a physically possible bumper separation between intact cars that cannot interpenetrate.

The actual collision has already happened at 2.09 s.

This is the conceptual limit of using stopping distance alone to study two moving vehicles.

For one car approaching a stationary obstacle, comparing its stopping position with the obstacle position is usually enough to determine whether contact occurs.

For two moving vehicles, the relative separation can become zero before either vehicle stops, even when their final positions would have been safely separated.

You must examine their positions at the same time, not just their final stopping distances.

06What should you take into the next problem?

For a reaction-time braking question, use this sequence:

1. Choose a positive direction and convert speeds to m/s if necessary.
2. Calculate how far the vehicle travels before braking takes effect.
3. Use uniform-acceleration equations for the braking stage, with the correct negative acceleration.
4. Combine the distances or positions only over the intervals where each equation is valid.
5. Check your result against the physical situation: obstacle position, stopping time, and whether a calculated root occurs before the vehicle stops.

Remember that the two-stage method assumes constant speed before braking and uniform deceleration afterwards. Real braking may involve changing grip, gradual brake engagement, or multiple acceleration stages.

The next useful question is how changing the starting speed affects both parts of the journey. Reaction distance increases directly with speed, but braking distance increases with speed squared when braking deceleration is fixed. That difference is explored in Does Doubling Speed Double Stopping Distance?.

For a stationary obstacle, calculate the distance before braking, then the distance during braking. For a moving obstacle, go one step further and track the separation throughout the motion.