How to Calculate Acceleration from Speed and Stopping Distance

A car is travelling at 20 m/s when its brakes are applied. It comes to rest 40 m later, but nobody records how long the braking takes. Can you still calculate its acceleration?

Yes. You only need the initial velocity, final velocity, and displacement.

For uniform acceleration, use:

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

Rearranging gives:

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

For an object that comes to a complete stop, the final velocity is zero. Therefore:

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

In this example, the acceleration is \(-5.0\ \mathrm{m\,s^{-2}}\). The negative sign tells us that the acceleration acts opposite to the car’s initial direction of travel.

This method is useful whenever an HSC Physics question gives you a speed and a stopping distance but doesn’t provide a stopping time.

01The calculation: understanding the equation

The equation \(v^2=u^2+2as\) is one of the constant-acceleration equations in physics.

Before using it, understand what each symbol represents.

There are two important conditions:

• The object must experience uniform acceleration.
• The displacement must be measured from the point where the initial velocity is recorded.

For a stopping-distance question, the final velocity is zero because the object has reached rest.

Substituting \(v=0\):

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

Rearranging:

\[ 2as=-u^2 \]

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

That’s the formula you need for a standard stopping-distance calculation.

Why don’t we need time?

You might remember the equation:

\[ v=u+at \]

This equation contains time, so it isn’t particularly helpful when the question doesn’t give you a time measurement.

The equation \(v^2=u^2+2as\) removes time by combining two other relationships:

\[ v=u+at \]

and

\[ s=\frac{u+v}{2}t \]

The second equation uses the average velocity during uniform acceleration.

Rearranging the first equation gives:

\[ t=\frac{v-u}{a} \]

Substituting into the displacement equation:

\[ s=\frac{u+v}{2}\times\frac{v-u}{a} \]

Using the difference of two squares:

\[ 2as=(u+v)(v-u) \]

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

Therefore:

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

Time disappears because the relationship already accounts for how velocity changes over the displacement.

02Getting the signs right

One of the most common mistakes in stopping-distance questions is getting the sign of acceleration wrong.

Let’s establish a sign convention.

Choose the initial direction of motion as positive.

For a car travelling to the right:

• Initial velocity is positive.
• Displacement to the stopping point is positive.
• Final velocity is zero.
• Acceleration is negative.

For example, suppose a vehicle travels at \(15\ \mathrm{m\,s^{-1}}\) and stops over 25 m.

\[ a=\frac{0^2-15^2}{2(25)} \]

\[ a=\frac{-225}{50} \]

\[ a=-4.5\ \mathrm{m\,s^{-2}} \]

The acceleration is negative because it points opposite to the vehicle’s initial motion.

The magnitude of the acceleration is \(4.5\ \mathrm{m\,s^{-2}}\).

This distinction matters:

• Acceleration is a vector, so it has direction.
• Deceleration describes a reduction in speed.
• A negative acceleration does not automatically mean an object is slowing down. Its effect depends on the direction of its velocity.

If you choose the opposite positive direction, the signs will change, but the physical acceleration will remain the same.

Speed and velocity are not interchangeable

Speed tells you how fast something moves.

Velocity tells you how fast it moves and in which direction.

This becomes important when an object stops and then reverses direction.

For example, a trolley initially moving right might have:

\[ u=+8\ \mathrm{m\,s^{-1}} \]

If it slows down, reaches zero velocity, and then moves left, its velocity becomes negative.

Its speed, however, remains positive.

When using the squared-velocity equation, remember that:

\[ v=\pm\sqrt{u^2+2as} \]

The positive and negative roots represent opposite directions of motion. They are not two different speeds.

For an ordinary stopping-distance calculation, the final velocity is exactly zero, so there is no ambiguity about which root to use.

For a problem involving reversal, you must consider which direction the object is travelling.

If the expression inside the square root is negative, there is no real velocity consistent with the assumed motion and constant-acceleration model.

03What does stopping distance actually measure?

Stopping distance is the displacement from the moment braking begins to the first moment the object reaches rest.

This is not always the same as the total distance travelled from the moment a driver notices a hazard.

There are two separate distances:

1. Reaction distance: the distance travelled while the driver responds.
2. Braking distance: the distance travelled after braking begins.

The total stopping distance is:

\[ s_{\text{total}}=s_{\text{reaction}}+s_{\text{braking}} \]

If the vehicle travels at constant speed during the reaction period:

\[ s_{\text{reaction}}=ut_r \]

where \(t_r\) is the reaction time.

The equation \(a=-u^2/(2s)\) uses the braking distance, not the total stopping distance.

This distinction is particularly important in questions involving reaction time. Using the total stopping distance in the braking equation would give an incorrect acceleration.

For a more detailed explanation of how physics distinguishes distance from displacement, see Distance and Displacement in HSC Physics Explained Clearly.

04Testing the method with an experiment

How could you measure acceleration if you didn’t have a sensor capable of measuring it directly?

You could measure the initial speed and stopping displacement, then calculate the acceleration using the equation.

A simple trolley experiment demonstrates the idea.

Setting up the apparatus

Imagine a dynamics trolley moving along a horizontal track.

A light string connects the rear of the trolley to a hanging mass through a pulley at the left end of the track. As the trolley moves to the right, the hanging mass rises, and the string provides an opposing force.

Near the starting point, a photogate measures the trolley’s initial speed. A flag of known length is attached to the trolley.

A motion sensor at the far end records the trolley’s position. This allows the first stopping point, where the trolley momentarily reaches zero velocity, to be identified.

What should the experiment show?

If the opposing net force is constant, Newton’s second law predicts constant acceleration, provided the total mass remains unchanged.

The trolley should slow down uniformly until it reaches its first stopping point.

The photogate measures the initial speed using:

\[ u=\frac{L}{\Delta t} \]

where:

• \(L\) is the length of the trolley’s flag.
• \(\Delta t\) is the time taken for the flag to pass through the photogate.

Suppose the flag is 0.10 m long and blocks the photogate for 0.050 s.

The initial speed is:

\[ u=\frac{0.10}{0.050} \]

\[ u=2.0\ \mathrm{m\,s^{-1}} \]

Now suppose the trolley reaches its first stopping point 0.75 m from the photogate.

Using the stopping-distance equation:

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

\[ a=-\frac{2.0^2}{2(0.75)} \]

\[ \boxed{a=-2.67\ \mathrm{m\,s^{-2}}} \]

The trolley’s acceleration is approximately \(2.67\ \mathrm{m\,s^{-2}}\) opposite its initial direction.

Comparing repeated measurements

The experiment could be repeated with different initial speeds while keeping the opposing force and trolley setup unchanged.

The following is an illustrative data set, not a record of an actual experiment.

The similar results support the constant-acceleration model.

Notice something interesting: increasing the initial speed increases the stopping distance.

In fact, rearranging the stopping equation gives:

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

If the acceleration magnitude stays constant, doubling the initial speed quadruples the stopping distance.

This is why a vehicle travelling twice as fast does not simply need twice the braking distance under the same constant-deceleration model.

What can the experiment not prove?

There is an important limitation.

Measuring only the initial speed and final stopping displacement cannot prove that acceleration was constant throughout the motion.

An object could experience stronger braking later in its journey and still reach rest at the same position.

The original measurements would not reveal that difference.

To improve the experiment, additional photogates could measure the trolley’s speed at several positions. Alternatively, a motion sensor could continuously record its position.

You could then calculate acceleration over separate intervals.

For uniform acceleration, a graph of \(v^2\) against displacement should be a straight line with gradient \(2a\).

If the gradient changes, the acceleration is not constant.

This is the key experimental limitation: the stopping-distance equation gives an exact result under the uniform-acceleration assumption, but the initial and final measurements alone cannot establish that assumption.

05Worked examples

06Questions and solutions

The following questions build from straightforward stopping calculations to more challenging problems involving reaction time, changing acceleration, and the interpretation of mathematical roots.

Question 1

A scooter travels along a straight, level path at an initial speed of \(15\ \mathrm{m\,s^{-1}}\). The rider applies the brakes, and the scooter comes to rest after travelling 25 m.

Assuming uniform acceleration, calculate the scooter’s acceleration.

Solution 1

The scooter’s acceleration is \(-4.5\ \mathrm{m\,s^{-2}}\), meaning it acts opposite to the scooter’s initial direction of motion.

Step 1: Identify the known quantities.

Taking the initial direction as positive:

\[ u=+15\ \mathrm{m\,s^{-1}} \]

\[ v=0 \]

\[ s=+25\ \mathrm{m} \]

Step 2: Apply the uniform-acceleration equation.

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

Step 3: Substitute the values.

\[ a=\frac{0^2-15^2}{2(25)} \]

\[ a=\frac{-225}{50} \]

\[ \boxed{a=-4.5\ \mathrm{m\,s^{-2}}} \]

The negative sign is important. Reporting only \(4.5\ \mathrm{m\,s^{-2}}\) would give the magnitude but not the direction of acceleration.

Question 2

A train travels along a straight track at \(30\ \mathrm{m\,s^{-1}}\). It brakes uniformly until its speed decreases to \(18\ \mathrm{m\,s^{-1}}\) after travelling 150 m.

Calculate:

a) The train’s acceleration.

b) The additional distance required for the train to come to rest if the same acceleration continues.

c) The total stopping distance measured from the point where braking first began.

Solution 2

The train’s acceleration is \(-1.92\ \mathrm{m\,s^{-2}}\), and its total stopping distance is approximately 234.4 m.

Step 1: Calculate the acceleration during the first 150 m.

Choose the direction of travel as positive.

\[ u=30\ \mathrm{m\,s^{-1}} \]

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

\[ s=150\ \mathrm{m} \]

Use:

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

Substitute:

\[ a=\frac{18^2-30^2}{2(150)} \]

\[ a=\frac{324-900}{300} \]

\[ \boxed{a=-1.92\ \mathrm{m\,s^{-2}}} \]

Step 2: Calculate the additional stopping distance.

The train is now travelling at \(18\ \mathrm{m\,s^{-1}}\).

Its final velocity will be zero.

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

\[ s=\frac{0^2-18^2}{2(-1.92)} \]

\[ s=\frac{-324}{-3.84} \]

\[ \boxed{s=84.375\ \mathrm{m}} \]

Step 3: Calculate the total stopping distance.

The train has already travelled 150 m before reaching 18 m/s.

Therefore:

\[ s_{\text{total}}=150+84.375 \]

\[ \boxed{s_{\text{total}}=234.4\ \mathrm{m}} \]

The trap here is treating the first 150 m as the stopping distance. The train still has considerable velocity at the end of that section, so its remaining stopping distance must also be included.

Question 3

A car travels towards an obstacle at \(24\ \mathrm{m\,s^{-1}}\). The obstacle is 62 m ahead when the driver first notices it.

The driver’s reaction time is 0.80 s. During this time, assume the car maintains constant speed.

Once the brakes are applied, the car experiences uniform acceleration of \(-6.0\ \mathrm{m\,s^{-2}}\).

Determine:

a) The reaction distance.

b) Whether the car can stop before reaching the obstacle.

c) The car’s speed when it reaches the obstacle, if it cannot stop in time.

Solution 3

The car travels 19.2 m during the reaction period. It cannot stop before the obstacle, and its speed at the obstacle is approximately \(7.9\ \mathrm{m\,s^{-1}}\).

Step 1: Calculate the reaction distance.

During the reaction period, the car maintains constant speed.

Use:

\[ s_{\text{reaction}}=ut_r \]

Substitute:

\[ s_{\text{reaction}}=24(0.80) \]

\[ \boxed{s_{\text{reaction}}=19.2\ \mathrm{m}} \]

Step 2: Calculate the distance available for braking.

The obstacle is 62 m away when the driver first notices it.

The remaining distance after the reaction period is:

\[ s_{\text{braking}}=62-19.2 \]

\[ \boxed{s_{\text{braking}}=42.8\ \mathrm{m}} \]

Step 3: Determine the distance required to stop.

At the moment braking begins:

\[ u=24\ \mathrm{m\,s^{-1}} \]

The final velocity for a complete stop is zero.

Using:

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

\[ s=\frac{0^2-24^2}{2(-6.0)} \]

\[ s=\frac{-576}{-12} \]

\[ \boxed{s_{\text{stop}}=48\ \mathrm{m}} \]

The available braking distance is only 42.8 m.

Therefore, the car cannot stop before reaching the obstacle.

Step 4: Calculate the speed at the obstacle.

The car has travelled 42.8 m under uniform braking.

Use:

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

Substitute:

\[ v^2=24^2+2(-6.0)(42.8) \]

\[ v^2=576-513.6 \]

\[ v^2=62.4 \]

Take the positive root because the car is still travelling forwards:

\[ v=\sqrt{62.4} \]

\[ \boxed{v=7.9\ \mathrm{m\,s^{-1}}} \]

The car reaches the obstacle with a speed of approximately \(7.9\ \mathrm{m\,s^{-1}}\).

The important reasoning is that reaction distance and braking distance are separate. Applying the braking equation to the entire 62 m would incorrectly assume that braking began as soon as the driver noticed the obstacle.

Question 4

A trolley initially travels at \(4.0\ \mathrm{m\,s^{-1}}\). It reaches its first stopping point after travelling 2.0 m.

A student calculates its acceleration using the uniform-acceleration equation and obtains:

\[ a=-4.0\ \mathrm{m\,s^{-2}} \]

The student then concludes that the trolley must have experienced constant acceleration throughout its motion.

Is this conclusion justified?

Construct an alternative motion involving two different constant accelerations, each acting over 1.0 m, that produces the same initial speed and final stopping position.

Finally, explain how an additional speed measurement at the 1.0 m mark could distinguish between the two models.

Solution 4

No. The value \(-4.0\ \mathrm{m\,s^{-2}}\) is consistent with uniform acceleration, but the initial speed and stopping distance alone do not prove that acceleration remained constant.

Step 1: Calculate the acceleration predicted by the uniform model.

The initial velocity is:

\[ u=4.0\ \mathrm{m\,s^{-1}} \]

The final velocity is zero, and the displacement is 2.0 m.

Therefore:

\[ a=\frac{0^2-4.0^2}{2(2.0)} \]

\[ \boxed{a=-4.0\ \mathrm{m\,s^{-2}}} \]

This is correct if acceleration is uniform.

However, we can construct another motion that gives exactly the same initial and final conditions.

Step 2: Consider an alternative first stage.

Suppose the trolley experiences an acceleration of:

\[ a_1=-2.0\ \mathrm{m\,s^{-2}} \]

over the first 1.0 m.

Using the uniform-acceleration equation:

\[ v_1^2=4.0^2+2(-2.0)(1.0) \]

\[ v_1^2=16-4 \]

\[ v_1^2=12 \]

Therefore:

\[ v_1=\sqrt{12} \]

\[ v_1=3.46\ \mathrm{m\,s^{-1}} \]

The trolley is still moving forwards at the 1.0 m mark.

Step 3: Consider the second stage.

Suppose the acceleration changes to:

\[ a_2=-6.0\ \mathrm{m\,s^{-2}} \]

over the next 1.0 m.

Using the velocity at the intermediate position:

\[ v_2^2=v_1^2+2a_2s \]

\[ v_2^2=12+2(-6.0)(1.0) \]

\[ v_2^2=12-12 \]

\[ \boxed{v_2=0} \]

The trolley stops after travelling exactly 2.0 m.

Both models therefore produce the same initial speed and stopping displacement.

Step 4: Determine how to distinguish the models.

Measure the trolley’s speed at the 1.0 m mark.

The uniform model predicts:

\[ v_1^2=4.0^2+2(-4.0)(1.0) \]

\[ v_1^2=8 \]

\[ \boxed{v_1=2.83\ \mathrm{m\,s^{-1}}} \]

The alternative two-stage model predicts:

\[ \boxed{v_1=3.46\ \mathrm{m\,s^{-1}}} \]

These predictions are different.

A photogate positioned at the 1.0 m mark could distinguish between them.

This question demonstrates an important limitation of the stopping-distance equation.

Knowing the initial speed and final stopping position allows you to calculate the acceleration that would produce the motion under a uniform-acceleration assumption. It does not establish that the actual acceleration was uniform.

Question 5

A trolley starts at position \(x=0\), travelling to the right with an initial velocity of \(+18\ \mathrm{m\,s^{-1}}\).

It experiences constant acceleration throughout its motion, including after it reaches zero velocity. The track is long enough for the trolley to reverse direction without hitting anything.

A fixed sensor is positioned at \(x=+16\ \mathrm{m}\).

The sensor records the trolley passing it twice. On both occasions, its measured speed is \(14\ \mathrm{m\,s^{-1}}\). The sensor measures speed only, not direction.

Determine:

a) The trolley’s acceleration.

b) The time of each crossing.

c) The distance from the starting position to the first stopping point.

d) The total distance travelled by the time the trolley passes the sensor for the second time.

Explain why both mathematical roots for the crossing times are physically valid, and why the sensor’s 16 m position is not the trolley’s stopping distance.

Solution 5

The trolley’s acceleration is \(-4.0\ \mathrm{m\,s^{-2}}\). It crosses the sensor at 1.0 s and 8.0 s, first stops 40.5 m from its starting position, and travels a total distance of 65.0 m by its second crossing.

The key is recognising that the sensor measures speed, not velocity. The trolley can pass the same position with the same speed while travelling in opposite directions.

Step 1: Calculate the acceleration.

The initial velocity is:

\[ u=+18\ \mathrm{m\,s^{-1}} \]

The displacement to the sensor is:

\[ s=+16\ \mathrm{m} \]

The speed at either crossing is:

\[ |v|=14\ \mathrm{m\,s^{-1}} \]

We can use the squared-velocity equation because it works with either possible velocity direction.

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

Substitute:

\[ 14^2=18^2+2a(16) \]

\[ 196=324+32a \]

\[ 32a=-128 \]

\[ \boxed{a=-4.0\ \mathrm{m\,s^{-2}}} \]

Notice that we did not need to decide whether the trolley was moving right or left at the sensor to calculate its acceleration.

Step 2: Calculate the two crossing times.

We now know the acceleration.

Use the displacement equation:

\[ s=ut+\frac{1}{2}at^2 \]

Substitute the known values:

\[ 16=18t+\frac{1}{2}(-4)t^2 \]

Simplify:

\[ 16=18t-2t^2 \]

Rearrange into quadratic form:

\[ 2t^2-18t+16=0 \]

Divide by 2:

\[ t^2-9t+8=0 \]

Factorise:

\[ (t-1)(t-8)=0 \]

Therefore:

\[ \boxed{t=1.0\ \mathrm{s}\quad\text{or}\quad t=8.0\ \mathrm{s}} \]

Both roots are positive, so both represent physically possible times after the trolley starts moving.

Why are there two answers?

At 1.0 s, the trolley is still travelling to the right.

Its velocity is:

\[ v=18+(-4)(1) \]

\[ \boxed{v=+14\ \mathrm{m\,s^{-1}}} \]

At 8.0 s:

\[ v=18+(-4)(8) \]

\[ \boxed{v=-14\ \mathrm{m\,s^{-1}}} \]

The trolley is now travelling to the left.

The sensor records a speed of 14 m/s on both occasions, but the velocity changes sign because the direction has reversed.

Step 3: Calculate the first stopping distance.

The trolley first stops when its velocity reaches zero.

Using:

\[ v=u+at \]

\[ 0=18-4t \]

\[ t=4.5\ \mathrm{s} \]

Now calculate its displacement at this time:

\[ s=ut+\frac{1}{2}at^2 \]

\[ s=18(4.5)+\frac{1}{2}(-4)(4.5)^2 \]

\[ s=81-40.5 \]

\[ \boxed{s_{\text{stop}}=40.5\ \mathrm{m}} \]

The trolley reaches its first stopping point 40.5 m to the right of its starting position.

Step 4: Calculate the total distance travelled by the second crossing.

The trolley travels:

• 40.5 m to reach its first stopping point.
• 24.5 m back towards the sensor.

The return distance is:

\[ 40.5-16=24.5\ \mathrm{m} \]

Therefore, its total distance travelled is:

\[ d_{\text{total}}=40.5+24.5 \]

\[ \boxed{d_{\text{total}}=65.0\ \mathrm{m}} \]

Step 5: Explain the two tempting mistakes.

The first mistake is treating the sensor position as the stopping distance.

Using the stopping formula with \(s=16\) m would give:

\[ a=-\frac{18^2}{2(16)} \]

\[ a=-10.125\ \mathrm{m\,s^{-2}} \]

This is incorrect because the trolley has not stopped at the sensor. Its speed is still 14 m/s.

The second mistake is substituting the total distance travelled, 65 m, into the squared-velocity equation.

That equation requires displacement, not total path length.

At the second crossing, the trolley is 16 m to the right of its starting position, even though it has travelled 65 m altogether.

The distinction between speed, velocity, distance, and displacement is what makes this problem challenging. Both roots are valid because the trolley passes the same position twice, once in each direction.

07Where to go next

Once you can calculate acceleration from speed and stopping distance, the next useful step is connecting these calculations to motion graphs.

In particular, understanding how velocity and acceleration change over time helps you recognise when uniform acceleration is a reasonable model and when it isn’t.

See Acceleration-Time Graphs: Area and Velocity Change to develop that connection.