Does Doubling Speed Double Stopping Distance? HSC Physics

At 20 m/s, a car brakes to a stop. Now imagine the same car travelling at 40 m/s, with exactly the same braking force and deceleration. Does it need twice the distance to stop?

No. It needs four times the braking distance.

There is an important catch, though. The distance travelled while the driver reacts to a hazard also counts towards the total stopping distance. That part only doubles when speed doubles. So total stopping distance generally increases by somewhere between two and four times, depending on how much of the original distance was spent reacting and how much was spent braking.

Let’s work out why, then look at how a simple trolley experiment can test the relationship.

01First, separate the two parts of stopping distance

Stopping distance is the total distance travelled between recognising a hazard and coming to a complete stop.

It has two components:

1. Thinking distance: The distance travelled while the driver recognises the hazard and reacts by applying the brakes.
2. Braking distance: The distance travelled after the brakes engage, until the vehicle stops.

Consider what happens during each stage.

The total stopping distance is therefore:

\[ d_{\text{total}}=d_{\text{thinking}}+d_{\text{braking}} \]

For the calculations in this article, we’ll assume:

• The vehicle travels along a straight, level road.
• Its speed remains constant during the driver’s reaction time.
• Its braking acceleration is uniform.
• The braking deceleration is the same when comparing different initial speeds.

These assumptions allow us to use the uniform acceleration equations.

If you need to refresh the difference between the distance travelled and displacement, see Distance and Displacement in HSC Physics Explained Clearly.

02Why does doubling speed quadruple braking distance?

Imagine two identical cars with identical brakes.

Car A travels at 10 m/s. Car B travels at 20 m/s.

Both begin braking with the same constant deceleration.

You might predict that Car B needs twice the distance because it is travelling twice as fast. The problem with that prediction is that it only considers how quickly the cars are travelling, not how long they need to slow down.

Let’s build the relationship properly.

Step 1: Understand the signs

Take the direction of travel as positive.

During braking:

• Initial velocity, \(u\), is positive.
• Final velocity, \(v\), is zero.
• Acceleration, \(a\), is negative.
• Displacement, \(s\), is positive because the vehicle continues forwards until it stops.

We can define \(b\) as the positive magnitude of the braking deceleration.

Therefore:

\[ a=-b \]

This distinction is useful. The acceleration is negative, but the deceleration magnitude is positive.

A negative acceleration does not mean the vehicle is travelling backwards. It means its velocity is decreasing in the positive direction.

Step 2: Use the uniform acceleration equation

The most useful equation for finding braking distance is:

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

At the instant the vehicle stops, \(v=0\). Substituting \(a=-b\):

\[ 0=u^2-2bs \]

Rearranging:

\[ \boxed{s=\frac{u^2}{2b}} \]

This is the key relationship.

For a fixed braking deceleration, braking distance is proportional to the square of initial speed:

\[ \boxed{s\propto u^2} \]

The speed is squared, not simply multiplied by the distance.

Step 3: Double the initial speed

Suppose the original speed is \(u\).

The original braking distance is:

\[ s_1=\frac{u^2}{2b} \]

Now double the speed to \(2u\):

\[ s_2=\frac{(2u)^2}{2b} \]

Expanding:

\[ s_2=\frac{4u^2}{2b} \]

Therefore:

\[ \boxed{s_2=4s_1} \]

Doubling speed quadruples braking distance.

Why does this make physical sense?

There is another way to picture the result.

With the same constant deceleration, doubling the initial speed also doubles the time needed to stop.

During braking, the average speed is half the initial speed because the vehicle slows uniformly to zero.

So, when the initial speed doubles:

• The braking time doubles.
• The average speed during braking doubles.

Distance equals average speed multiplied by time.

Doubling both quantities gives four times the distance.

This is a useful mental check when you’re unsure whether a braking-distance calculation should involve \(u\) or \(u^2\).

03Testing the relationship with a trolley experiment

How could we check whether braking distance really increases with the square of initial speed?

A simple school-laboratory investigation can test the prediction using a trolley, a photogate, and a controlled braking surface.

The experimental setup

Imagine a horizontal track.

A spring launcher at the left sends a trolley towards a braking strip. Immediately before the strip, a photogate measures the trolley’s initial speed.

The trolley has a spring-loaded brake pad underneath it. The pad presses against the same rough strip during each trial, producing an approximately constant frictional force.

A ruler runs alongside the braking strip so that the distance travelled between entering the strip and coming to rest can be measured.

The investigation needs to change only one important variable: the trolley’s initial speed.

Keep the following conditions constant:

• The mass of the trolley.
• The braking pad and its contact pressure.
• The braking surface.
• The horizontal track alignment.
• The conditions of the surface between trials.

Vary the launch setting to produce different initial speeds.

The photogate measures the actual speed at the beginning of braking. This is more reliable than assuming that a particular launcher setting produces a particular speed.

What should the experiment show?

If the braking deceleration is constant, our equation predicts:

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

Therefore, a graph of braking distance against speed squared should be a straight line through the origin.

Consider the following illustrative data.

These are illustrative model-based readings, not measurements from an actual experiment.

Notice what happens when the speed doubles:

• From 0.50 to 1.00 m/s, braking distance increases from approximately 0.063 m to 0.250 m.
• From 1.00 to 2.00 m/s, braking distance increases from 0.250 m to 1.000 m.

In both comparisons, the braking distance increases by a factor of four.

What can we calculate from the results?

We can use one reading to estimate the braking deceleration.

Take the trial with an initial speed of 1.50 m/s and a braking distance of 0.563 m.

Starting with:

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

Rearrange to make \(b\) the subject:

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

Substitute:

\[ b=\frac{1.50^2}{2(0.563)} \]

\[ b=\frac{2.25}{1.126} \]

\[ \boxed{b\approx2.00\ \text{m/s}^2} \]

The estimated braking deceleration is approximately 2.00 m/s².

The other readings give approximately the same value. This is consistent with the constant-deceleration model.

Why does the experimental design matter?

Suppose the original experiment simply released the trolley from different heights on a ramp and measured the distance from the release point to where the trolley stopped.

That design has a problem.

The measured distance includes the trolley’s motion before braking begins. It does not isolate braking distance. Also, the release height does not directly measure the speed at the start of braking.

The revised design addresses both issues:

• The photogate measures the actual initial speed.
• The ruler starts at the beginning of the braking strip.
• The same trolley, pad, and surface are used for each trial.

Even then, the results only support the model within the accuracy of the experiment. Friction can vary slightly between trials, so real measurements will not necessarily follow a perfect square relationship.

The important conclusion is that the square relationship follows from the assumption of constant braking deceleration. The experiment tests how well that assumption describes the trolley’s motion.

06When does the four-times rule apply?

The relationship is worth summarising because it is an easy place to lose marks through incorrect assumptions.

For the same constant braking deceleration:

Here, \(T\) is the original thinking distance and \(B\) is the original braking distance.

The total stopping distance can be expressed as:

\[ D(u)=ut_r+\frac{u^2}{2b} \]

At twice the speed:

\[ D(2u)=2ut_r+\frac{4u^2}{2b} \]

The total distance therefore does not follow a simple square relationship when reaction time is included.

For positive reaction time and braking distance, doubling speed increases total stopping distance by a factor between two and four.

There are also other situations where the four-times rule cannot be applied directly.

Different braking decelerations: If one vehicle has a different braking deceleration, the comparison changes. Remember that braking distance is inversely proportional to the magnitude of deceleration.

Changing acceleration: If the braking acceleration changes during the motion, you cannot automatically use one constant value for the entire stopping process. Divide the motion into separate intervals and calculate each interval independently.

Different reaction times: If the drivers have different reaction times, their thinking distances will also differ. Comparing only their initial speeds is not enough.

The square relationship is a result of a particular physical model, not a universal rule that applies to every stopping situation.

07Questions and solutions

Question 1

A trolley travels along a horizontal track at 10.0 m/s. It experiences a constant acceleration of -2.5 m/s² until it comes to rest.

Calculate:

a) The time taken to stop.

b) The braking distance.

Solution 1

The trolley stops after 4.0 seconds and travels 20.0 metres while braking.

Step 1: Identify the known quantities.

\[ u=10.0\ \text{m/s} \]

\[ v=0 \]

\[ a=-2.5\ \text{m/s}^2 \]

Take the initial direction of travel as positive.

Step 2: Calculate stopping time.

Use:

\[ v=u+at \]

\[ 0=10.0-2.5t \]

\[ t=\frac{10.0}{2.5} \]

\[ \boxed{t=4.0\ \text{s}} \]

Step 3: Calculate braking distance.

Use:

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

\[ 0=10.0^2+2(-2.5)s \]

\[ 0=100-5s \]

\[ \boxed{s=20.0\ \text{m}} \]

The positive displacement makes sense because the trolley continues forwards while slowing down.

Question 2

A car is travelling at 22.0 m/s. The driver has a reaction time of 0.70 seconds, after which the car brakes with a constant acceleration of -5.0 m/s².

a) Calculate the thinking distance, braking distance, and total stopping distance.

b) The car now travels at 33.0 m/s. Its reaction time and braking deceleration remain unchanged. Calculate its new total stopping distance.

c) A student claims that increasing speed by 50% must increase total stopping distance by 50%. Evaluate this claim using your results.

Solution 2

The total stopping distances are 63.8 metres at 22.0 m/s and 132.0 metres at 33.0 m/s. The student’s claim is incorrect.

Step 1: Calculate the thinking distance at 22.0 m/s.

\[ d_{\text{thinking}}=ut_r \]

\[ =22.0(0.70) \]

\[ \boxed{d_{\text{thinking}}=15.4\ \text{m}} \]

Step 2: Calculate the braking distance.

The final velocity is zero.

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

\[ 0=22.0^2+2(-5.0)s \]

\[ 0=484-10s \]

\[ \boxed{d_{\text{braking}}=48.4\ \text{m}} \]

Step 3: Find total stopping distance.

\[ d_{\text{total}}=15.4+48.4 \]

\[ \boxed{d_{\text{total}}=63.8\ \text{m}} \]

Step 4: Repeat at 33.0 m/s.

Thinking distance:

\[ d_{\text{thinking}}=33.0(0.70) \]

\[ \boxed{d_{\text{thinking}}=23.1\ \text{m}} \]

Braking distance:

\[ 0=33.0^2+2(-5.0)s \]

\[ 0=1089-10s \]

\[ \boxed{d_{\text{braking}}=108.9\ \text{m}} \]

Total:

\[ d_{\text{total}}=23.1+108.9 \]

\[ \boxed{d_{\text{total}}=132.0\ \text{m}} \]

Step 5: Evaluate the student’s claim.

Compare the total distances:

\[ \text{Ratio}=\frac{132.0}{63.8} \]

\[ \boxed{\text{Ratio}\approx2.07} \]

The speed increased by a factor of 1.5, but total stopping distance increased by approximately 2.07 times.

The student’s mistake is assuming that the entire stopping distance is directly proportional to speed.

Thinking distance increases by a factor of 1.5, but braking distance increases by:

\[ 1.5^2=2.25 \]

The combined result is therefore not a 50% increase.

Question 3

A test trolley enters a braking track at 16.0 m/s.

Its acceleration changes during braking:

• During the first stage, its acceleration is -3.0 m/s² until its speed decreases to 7.0 m/s.
• During the second stage, its acceleration is -1.4 m/s² until it stops.

Calculate:

a) The time taken during each stage.

b) The distance travelled during each stage.

c) The total braking distance and total braking time.

A student suggests averaging the two accelerations and treating the entire motion as one uniformly accelerated interval. Explain why this method does not give the correct braking distance.

Solution 3

The trolley travels a total braking distance of 52.0 metres in 8.0 seconds.

The two stages must be calculated separately because the acceleration changes during the motion.

Step 1: Calculate the time taken during the first stage.

The initial velocity is 16.0 m/s and the final velocity is 7.0 m/s.

Use:

\[ v=u+at \]

\[ 7.0=16.0-3.0t_1 \]

\[ t_1=\frac{9.0}{3.0} \]

\[ \boxed{t_1=3.0\ \text{s}} \]

Step 2: Calculate the first-stage distance.

Use:

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

\[ 7.0^2=16.0^2+2(-3.0)s_1 \]

\[ 49=256-6s_1 \]

\[ s_1=\frac{207}{6} \]

\[ \boxed{s_1=34.5\ \text{m}} \]

Step 3: Calculate the time taken during the second stage.

The new initial velocity is 7.0 m/s, and the final velocity is zero.

\[ 0=7.0-1.4t_2 \]

\[ \boxed{t_2=5.0\ \text{s}} \]

Step 4: Calculate the second-stage distance.

\[ 0^2=7.0^2+2(-1.4)s_2 \]

\[ 0=49-2.8s_2 \]

\[ \boxed{s_2=17.5\ \text{m}} \]

Step 5: Calculate the totals.

Total braking distance:

\[ s_{\text{total}}=34.5+17.5 \]

\[ \boxed{s_{\text{total}}=52.0\ \text{m}} \]

Total braking time:

\[ t_{\text{total}}=3.0+5.0 \]

\[ \boxed{t_{\text{total}}=8.0\ \text{s}} \]

Step 6: Explain why averaging the accelerations is not sufficient.

The student’s arithmetic average would be:

\[ a_{\text{average}}=\frac{-3.0+(-1.4)}{2} \]

\[ =-2.2\ \text{m/s}^2 \]

This is incorrect because the two accelerations do not act for equal amounts of time.

The time-weighted average acceleration is actually:

\[ a_{\text{average}}= \frac{(-3.0)(3.0)+(-1.4)(5.0)}{8.0} \]

\[ =\frac{-16}{8} \]

\[ =-2.0\ \text{m/s}^2 \]

Even this correct average acceleration does not provide the correct displacement if we pretend that the acceleration was constant throughout.

A constant acceleration of -2.0 m/s² over eight seconds would produce a distance of 64.0 metres, not 52.0 metres.

Why?

Because acceleration determines the shape of the velocity-time graph. The actual trolley loses speed quickly at first, then spends longer moving at relatively low speeds.

Its displacement depends on the area under the actual velocity-time graph, not just the average acceleration.

Question 4

A trolley passes a photogate at 12.0 m/s and immediately begins braking with constant acceleration.

A point P is located 18.0 metres beyond the photogate. The trolley comes to rest 6.0 metres beyond point P.

Calculate:

a) The trolley’s acceleration.

b) Its velocity when it passes point P.

c) The time taken to reach point P.

d) The total time taken to stop.

Explain why the negative mathematical root for velocity at point P must be rejected.

Solution 4

The acceleration is -3.0 m/s², the velocity at P is +6.0 m/s, and the trolley takes 4.0 seconds to stop.

Step 1: Calculate the total braking distance.

The trolley travels 18.0 metres to P and another 6.0 metres before stopping.

\[ s_{\text{total}}=18.0+6.0 \]

\[ \boxed{s_{\text{total}}=24.0\ \text{m}} \]

Step 2: Find the acceleration.

Use:

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

At the final position, \(v=0\).

\[ 0=12.0^2+2a(24.0) \]

\[ 0=144+48a \]

\[ a=-\frac{144}{48} \]

\[ \boxed{a=-3.0\ \text{m/s}^2} \]

The negative sign indicates that acceleration acts opposite to the initial direction of travel.

Step 3: Calculate velocity at P.

The displacement from the photogate to P is 18.0 metres.

Use:

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

\[ v_P^2=12.0^2+2(-3.0)(18.0) \]

\[ v_P^2=144-108 \]

\[ v_P^2=36 \]

Taking the square root:

\[ v_P=\pm6.0\ \text{m/s} \]

We must choose the positive root:

\[ \boxed{v_P=+6.0\ \text{m/s}} \]

Why reject the negative root?

The equation gives the square of velocity, so both positive and negative mathematical roots are possible.

However, point P is reached before the trolley stops. The trolley is still travelling forwards, so its velocity must be positive.

A velocity of -6.0 m/s would mean that the trolley had reversed direction. That contradicts the information that it comes to rest beyond P.

Step 4: Calculate the time to reach P.

Use:

\[ v=u+at \]

\[ 6.0=12.0-3.0t_P \]

\[ \boxed{t_P=2.0\ \text{s}} \]

Step 5: Calculate the remaining stopping time.

From P, the trolley slows from 6.0 m/s to zero.

\[ 0=6.0-3.0t_2 \]

\[ \boxed{t_2=2.0\ \text{s}} \]

Total time:

\[ t_{\text{total}}=2.0+2.0 \]

\[ \boxed{t_{\text{total}}=4.0\ \text{s}} \]

The trolley reaches P after two seconds and stops after four seconds.

The important lesson is that the sign of a velocity must agree with the physical direction of motion. Squaring an equation can hide that information, so always interpret the roots.

Question 5

Two identical test vehicles are investigated using the same two-stage braking program.

Vehicle A begins at 12.0 m/s, while Vehicle B begins at 24.0 m/s.

Both vehicles have a response delay of 0.25 seconds before their brakes engage. During this delay, they continue at constant speed.

Once braking begins, the following conditions apply to both vehicles:

• First stage: constant acceleration of -4.0 m/s² until the speed reaches 6.0 m/s.
• Second stage: constant acceleration of -1.5 m/s² until the vehicle stops.

Calculate:

a) The total stopping distance for each vehicle, including the response delay.

b) The ratio of their total stopping distances.

c) Whether the ratio would remain the same if the response delay were removed.

A student argues that Vehicle B must travel exactly four times as far as Vehicle A because its initial speed is doubled.

Explain why this prediction is incorrect, even though both vehicles follow the same braking program.

Solution 5

Vehicle A has a total stopping distance of 28.5 metres. Vehicle B has a total stopping distance of 85.5 metres.

Their total stopping-distance ratio is exactly 3:1, not 4:1.

The key is to calculate the response delay and both braking stages separately.

Step 1: Calculate Vehicle A’s response distance.

Vehicle A travels at 12.0 m/s for 0.25 seconds before braking.

\[ d_{\text{response,A}}=ut \]

\[ =12.0(0.25) \]

\[ \boxed{d_{\text{response,A}}=3.0\ \text{m}} \]

Step 2: Calculate Vehicle A’s first braking stage.

The velocity decreases from 12.0 m/s to 6.0 m/s.

First, calculate the time:

\[ 6.0=12.0-4.0t_1 \]

\[ \boxed{t_1=1.5\ \text{s}} \]

Now calculate the distance:

\[ 6.0^2=12.0^2+2(-4.0)s_1 \]

\[ 36=144-8s_1 \]

\[ \boxed{s_1=13.5\ \text{m}} \]

Step 3: Calculate Vehicle A’s second braking stage.

The velocity decreases from 6.0 m/s to zero.

Time:

\[ 0=6.0-1.5t_2 \]

\[ \boxed{t_2=4.0\ \text{s}} \]

Distance:

\[ 0^2=6.0^2+2(-1.5)s_2 \]

\[ 0=36-3s_2 \]

\[ \boxed{s_2=12.0\ \text{m}} \]

Step 4: Find Vehicle A’s total stopping distance.

Add all three distances:

\[ d_A=3.0+13.5+12.0 \]

\[ \boxed{d_A=28.5\ \text{m}} \]

Step 5: Calculate Vehicle B’s response distance.

Vehicle B travels at 24.0 m/s during the same 0.25-second delay.

\[ d_{\text{response,B}}=24.0(0.25) \]

\[ \boxed{d_{\text{response,B}}=6.0\ \text{m}} \]

Step 6: Calculate Vehicle B’s first braking stage.

The velocity decreases from 24.0 m/s to 6.0 m/s.

Time:

\[ 6.0=24.0-4.0t_1 \]

\[ \boxed{t_1=4.5\ \text{s}} \]

Distance:

\[ 6.0^2=24.0^2+2(-4.0)s_1 \]

\[ 36=576-8s_1 \]

\[ \boxed{s_1=67.5\ \text{m}} \]

Step 7: Calculate Vehicle B’s second braking stage.

This stage starts at the same velocity of 6.0 m/s and has the same acceleration of -1.5 m/s².

Therefore, the calculation is identical to Vehicle A’s second stage.

\[ \boxed{s_2=12.0\ \text{m}} \]

Step 8: Calculate Vehicle B’s total stopping distance.

\[ d_B=6.0+67.5+12.0 \]

\[ \boxed{d_B=85.5\ \text{m}} \]

Step 9: Compare the total distances.

\[ \text{Ratio}=\frac{85.5}{28.5} \]

\[ \boxed{\text{Ratio}=3.00} \]

Vehicle B travels exactly three times the total stopping distance of Vehicle A.

Step 10: Investigate what happens if the response delay is removed.

Without the response delay, only the braking distances remain.

Vehicle A:

\[ d_{A,\text{braking}}=13.5+12.0 \]

\[ =25.5\ \text{m} \]

Vehicle B:

\[ d_{B,\text{braking}}=67.5+12.0 \]

\[ =79.5\ \text{m} \]

The new ratio is:

\[ \frac{79.5}{25.5}\approx3.12 \]

Therefore, removing the response delay changes the ratio from 3.00 to approximately 3.12.

Why doesn’t the four-times rule apply?

The tempting approach is to assume that doubling the initial speed must quadruple braking distance.

That conclusion is only valid when the same constant deceleration acts throughout the entire braking interval.

Here, acceleration changes when the speed reaches 6.0 m/s.

Both vehicles therefore have exactly the same final braking stage: a reduction from 6.0 m/s to zero at -1.5 m/s². This stage contributes 12.0 metres to each vehicle’s braking distance.

That common 12.0-metre contribution does not quadruple when the initial speed doubles.

The response distance also only doubles because it depends linearly on speed.

The square relationship has not disappeared. Its original conditions simply no longer describe the entire motion.

This is the important distinction: the same braking program does not necessarily mean one constant braking acceleration throughout the journey.

The next useful concept is the velocity-time graph. Its gradient represents acceleration, while the area underneath it represents displacement. For a vehicle with changing braking acceleration, the graph lets you see why each stage must be treated separately and why simply averaging accelerations can produce the wrong stopping distance.