Motion With Constant Acceleration (Examples, Exercises, Solutions)

An abstraction often used in physics is motion with constant acceleration. This is a good approximation for many different situations: free fall over small distances or in low-density atmospheres, full braking in car traffic, an object sliding down an inclined plane, etc … The mathematics behind this special case is relatively simple. Assume the object that is subject to the constant acceleration a (in m/s²) initially has a velocity v(0) (in m/s). Since the velocity is the integral of the acceleration function, the object’s velocity after time t (in s) is simply:

1) v(t) = v(0) + a · t

For example, if a car initially goes v(0) = 20 m/s and brakes with a constant a = -10 m/s², which is a realistic value for asphalt, its velocity after a time t is:

v(t) = 20 – 10 · t

After t = 1 second, the car’s speed has decreased to v(1) = 20 – 10 · 1 = 10 m/s and after t = 2 seconds the car has come to a halt: v(2) = 20 – 10 · 2 = 0 m/s. As you can see, it’s all pretty straight-forward. Note that the negative acceleration (also called deceleration) has led the velocity to decrease over time. In a similar manner, a positive acceleration will cause the speed to go up. You can read more on acceleration in this blog post.

What about the distance x (in m) the object covers? We have to integrate the velocity function to find the appropriate formula. The covered distance after time t is:

2) x(t) = v(0) · t + 0.5 · a · t²

While that looks a lot more complicated, it is really just as straight-forward. Let’s go back to the car that initially has a speed of v(0) = 20 m/s and brakes with a constant a = -10 m/s². In this case the above formula becomes:

x(t) = 20 · t – 0.5 · 10 · t²

After t = 1 second, the car has traveled x(1) = 20 · 1 – 0.5 · 10 · 1² = 15 meters. By the time it comes to a halt at t = 2 seconds, it moved x(2) = 20 · 2 – 0.5 · 10 · 2² = 20 meters. Note that we don’t have to use the time as a variable. There’s a way to eliminate it. We could solve equation 1) for t and insert the resulting expression into equation 2). This leads to a formula connecting the velocity v and distance x.

3) Constant acceleration_html_b85f3ec

Solved for x it looks like this:

3)’ Constant acceleration_html_m23bb2bb3

It’s a very useful formula that you should keep in mind. Suppose a tram accelerates at a constant a = 1.3 m/s², which is also a realistic value, from rest (v(0) = 0 m/s). What distance does it need to go to full speed v = 10 m/s? Using equation 3)’ we can easily calculate this:

Constant acceleration_html_m11de6604


Here are a few exercises and solutions using the equations 1), 2) and 3).

1. During free fall (air resistance neglected) an object accelerates with about a = 10 m/s. Suppose the object is dropped, that is, it is initially at rest (v(0) = 0 m/s).

a) What is its speed after t = 3 seconds?
b) What distance has it traveled after t = 3 seconds?
c) Suppose we drop the object from a tower that is x = 20 meters tall. At what speed will it impact the ground?
d) How long does the drop take?

Hint: in exercise d) solve equation 1) for t and insert the result from c)

2. During the reentry of space crafts accelerations can be as high as a = -70 m/s². Suppose the space craft initially moves with v(0) = 6000 m/s.

a) What’s the speed and covered distance after t = 10 seconds?
b) How long will it take the space craft to half its initial velocity?
c) What distance will it travel during this time?

3. An investigator arrives at the scene of a car crash. From the skid marks he deduces that it took the car a distance x = 55 meters to come to a halt. Assume full braking (a = -10 m/s²). Was the car initially above the speed limit of 30 m/s?


Solutions to the exercises:

Exercise 1

a) 30 m/s
b) 45 m
c) 20 m/s
d) 2 s

Exercise 2

a) 5,300 m/s and 56,500 m
b) 42.9 s (rounded)
c) 192,860 m (rounded)

Exercise 3

Yes (he was initially going 33.2 m/s)


To learn the basic math you need to succeed in physics, check out the e-book “Algebra – The Very Basics”. For an informal introduction to physics, check out the e-book “Physics! In Quantities and Examples”. Both are available at low prices and exclusively for Kindle.


A Brief Look At Car-Following Models

Recently I posted a short introduction to recurrence relations – what they are and how they can be used for mathematical modeling. This post expands on the topic as car-following models are a nice example of recurrence relations applied to the real-world.

Suppose a car is traveling on the road at the speed u(t) at time t. Another car approaches this car from behind and starts following it. Obviously the driver of the car that is following cannot choose his speed freely. Rather, his speed v(t) at time t will be a result of whatever the driver in the leading car is doing.

The most basic car-following model assumes that the acceleration a(t) at time t of the follower is determined by the difference in speeds. If the leader is faster than the follower, the follower accelerates. If the leader is slower than the follower, the follower decelerates. The follower assumes a constant speed if there’s no speed difference. In mathematical form, this statement looks like this:

a(t) = λ * (u(t) – v(t))

The factor λ (sensitivity) determines how strongly the follower accelerates in response to a speed difference. To be more specific: it is the acceleration that results from a speed difference of one unit.


Before we go on: how is this a recurrence relation? In a recurrence relation we determine a quantity from its values at an earlier time. This seems to be missing here. But remember that the acceleration is given by:

a(t) = (v(t+h) – v(t)) / h

with h being a time span. Inserted into the above car-following equation, we can see that it indeed implies a recurrence relation.


Our model is still very crude. Here’s the biggest problem: The response of the driver is instantaneous. He picks up the speed difference at time t and turns this information into an acceleration also at time t. But more realistically, there will be a time lag. His response at time t will be a result of the speed difference at an earlier time t – Λ, with Λ being the reaction time.

a(t) = λ * (u(t – Λ) – v(t – Λ))

The reaction time is usually in the order of one second and consist of the time needed to process the information as well as the time it takes to move the muscles and press the pedal. There are several things we can do to make the model even more realistic. First of all, studies show that the speed difference is not the only factor. The distance d(t) between the leader and follower also plays an important role. The smaller it is, the stronger the follower will react. We can take this into account by putting the distance in the denominator:

a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ))

You can also interpret this as making the sensitivity distance-dependent. There’s still one adjustment we need to make. The above model allows any value of acceleration, but we know that we can only reach certain maximum values in a car. Let’s symbolize the maximum acceleration by a(acc) and the maximum deceleration by a(dec). The latter will be a number smaller than zero since deceleration is by definition negative acceleration. We can write:

a(t) = a(acc) if (λ / d(t)) * (u(t – Λ) – v(t – Λ)) > a(acc)
a(t) = a(dec) if (λ / d(t)) * (u(t – Λ) – v(t – Λ)) < a(dec)
a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ)) else

It probably looks simpler using an if-statement:

a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ))

IF a(t) > a(acc) THEN
a(t) = a(acc)
ELSEIF a(t) < a(dec) THEN
a(t) = a(dec)

This model already catches a lot of nuances of car traffic. I hope I was able to give you some  insight into what car-following models are and how you can fine-tune them to satisfy certain conditions.

What is Mass? A Short and Simple Explanation

Mass is such a fundamental property of matter that it is hard to define without drifting into philosophical realms. Newton’s Second Law provides a great way to understand mass from a physical point of view. The law states that force F (in N) is the product of mass m (in kg) and acceleration a (in m/s²):

F = m · a

So according to this, mass is a measure of an object’s resistance to a change in speed. If the mass is small, a small force is sufficient to produce a noticeable acceleration. However, much more force is necessary to produce the same acceleration for a massive object.

Another way of looking at mass is provided by Newton’s Law of Gravitation. Newton found that the attracting gravitational force between two objects is proportional to the product of their masses m and M:

F ~ m · M

So additionally to creating resistance to changes in state of motion, mass is also the source of gravitational attraction. It seems obvious that in both cases we are talking about the same quantity. But is this actually the case? Is the inertial mass, the mass responsible for opposing changes in velocity, really the same as the gravitational mass, that gives rise to gravity?

This question has led to heated debates among physicist for centuries. All experiments conducted so far, with ever increasing accuracy, have shown that indeed the inertial mass is identical to the gravitational mass. Today, almost all physicists have accepted this equivalence as reality.

The SI unit of mass is kilograms. Ever since 1889, one kilogram has been defined as the mass of the international prototype kilogram (IPK) that is stored in the International Bureau of Weights and Measures in Paris. However, during the 24th General Conference on Weights and Measures that took place in 2011, physicists have agreed to redefine this unit by connecting it to the Planck constant.

Other units that are commonly used for mass are grams (1/1000 of a kilogram), the pound (equal to about 0.45 kilograms) and the tonne (equal to 1000 kilograms). For atoms and molecules scientists use the atomic mass unit u. One u is equivalent to 1.66 · 10-27 kg, which is roughly the mass of a neutron or proton.

(This was an excerpt from Physics! In Quantities and Examples)

Released Today for Kindle: Physics! In Quantities and Examples

I finally finished and released my new ebook … took me longer than usual because I always kept finding new interesting topics while researching. Here’s the blurb, link and TOC:

This book is a concept-focused and informal introduction to the field of physics that can be enjoyed without any prior knowledge. Step by step and using many examples and illustrations, the most important quantities in physics are gently explained. From length and mass, over energy and power, all the way to voltage and magnetic flux. The mathematics in the book is strictly limited to basic high school algebra to allow anyone to get in and to assure that the focus always remains on the core physical concepts.

(Click cover to get to the Amazon Product Page)


Table of Contents:

(Introduction, From the Smallest to the Largest, Wavelength)

(Introduction, Mass versus Weight, From the Smallest to the Largest, Mass Defect and Einstein, Jeans Mass)

Speed / Velocity
(Introduction, From the Smallest to the Largest, Faster than Light, Speed of Sound for all Purposes)

(Introduction, From the Smallest to the Largest, Car Performance, Accident Investigation)

(Introduction, Thrust and the Space Shuttle, Force of Light and Solar Sails, MoND and Dark Matter, Artificial Gravity and Centrifugal Force, Why do Airplanes Fly?)

(Introduction, Surface Area and Heat, Projected Area and Planetary Temperature)

(Introduction, From the Smallest to the Largest, Hydraulic Press, Air Pressure, Magdeburg Hemispheres)

(Introduction, Poisson’s Ratio)

(Introduction, From the Smallest to the Largest, Bulk Density, Water Anomaly, More Densities)

(Introduction, From the Smallest to the Largest, Thermal Expansion, Boiling, Evaporation is Cool, Why Blankets Work, Cricket Temperature)

(Introduction, Impact Speed, Ice Skating, Dear Radioactive Ladies and Gentlemen!, Space Shuttle Reentry, Radiation Exposure)

(Introduction, From the Smallest to the Largest, Space Shuttle Launch and Sound Suppression)

(Introduction, Inverse Square Law, Absorption)

(Introduction, Perfectly Inelastic Collisions, Recoil, Hollywood and Physics, Force Revisited)

Frequency / Period
(Introduction, Heart Beat, Neutron Stars, Gravitational Redshift)

Rotational Motion
(Extended Introduction, Moment of Inertia – The Concept, Moment of Inertia – The Computation, Conservation of Angular Momentum)

(Extended Introduction, Stewart-Tolman Effect, Piezoelectricity, Lightning)

(Extended Introduction, Lorentz Force, Mass Spectrometers, MHD Generators, Earth’s Magnetic Field)

Scalar and Vector Quantities
Measuring Quantities
Unit Conversion
Unit Prefixes
Copyright and Disclaimer

As always, I discounted the book in countries with a low GDP because I think that education should be accessible for all people. Enjoy!

Another Home Experiment – Wind Speed and Sound Level

Recently I told you about my home experiment regarding impact speed and sound level. I did another experiment with my sound level meter, this time I was interested in finding out how the sound level varies with the wind speed. So I took my anemometer (yep, that’s a thing) to measure the wind speed and at the same time noted the sound level. I collected some data points and plotted them. Here’s the result:


As you can see the fit is not that bad (the adjusted r-square is 0.91).

So the sound level grows with the wind velocity to the power of 0.22, meaning that if the wind speed increases by a factor of twenty-five, the sound level doubles. According to the empirical formula, the noise from the wind inside a category 1 and 2 hurricane is comparable to the sound level at a rock concert. This is of course assuming that the formula holds true past the 12 m/s range over which it was determined (which is not necessarily the case, but for now the best guess).

Home Experiment – Impact Speed and Sound Level

A while ago I got my hands on a sound level meter and pondered what to do with it. Sound level versus distance from source? Too boring, there’s already a formula for that (see here: Intensity: How Much Power Will Burst Your Eardrums?). What I noticed though is that I’ve never seen a formula relating impact height or speed to sound level, that seemed interesting. So I bought a small wooden sphere at a local store and dropped it from various heights, at each impact recording the maximum sound level. I dropped the sphere from 8 different heights and to reduce the effect of random fluctuations 20 times from each height. So in total I collected 160 data points. I’m not so sure if my neighbors were happy about that.

I calculated the impact speed v from the drop height h using the common v = sqrt (2 * g * h). As you might know, this formula neglects air resistance. However, I’m not concerned about that. The wooden sphere was small and massive and only dropped from heights below about 1 ft. The computed impact speed shouldn’t be off by more than a few percent.

Here’s the resulting plot of impact speed versus sound level (in decibels):

Impact Speed Sound Level Decibel

The fit turned out to be fantastic and implies that if you increase the impact speed by a factor of five, the sound level doubles. What’s the point of this? I don’t know, but it’s a neat graph and that’s good enough for me.

Acceleration – A Short and Simple Explanation

The three basic quantities used in kinematics are distance, velocity and acceleration. Let’s first look at velocity before moving on to the main topic. The velocity is simply the rate of change in distance. If we cover the distance d in a time span t, than the average velocity during this interval is:

v = d / t

So if we drive d = 800 meters in t = 40 seconds, the average speed is v = 800 meters / 40 seconds = 20 m/s. No surprise here. Note that there are many different units commonly used for velocity: kilometers per hour, feet per second, miles per hour, etc … The SI unit is m/s, so unless otherwise stated, you have to input the velocity in m/s into a formula to get a correct result.

Acceleration is also defined as the rate of change, but this time with respect to velocity. If the velocity changes by the amount v in a time span t, the average acceleration is:

a = v / t

For example, my beloved Mercedes C-180 Compressor can go from 0 to 100 kilometers per hour (or 27.8 meters per second) in about 9 seconds. So the average acceleration during this time is:

a = 27.8 meters per second / 9 seconds = 3.1 m/s²

Is that a lot? Obviously we should know some reference values to be able to judge acceleration.

The one value you should know is: g = 9.81 m/s². This is the acceleration experienced in free fall. And you can take the word “experienced” literally because unlike velocity, we really do feel acceleration. Our inner ear system contains structures that enable us to perceive it. Often times acceleration is compared to this value because it provides a meaningful and easily relatable reference value.

So the acceleration in the Mercedes C-180 Compressor is not quite as thrilling as free fall, it only accelerates with about 3.1 / 9.81 = 0.32 g. How much higher can it go for production cars? Well, meet the Bugatti Veyron Super Sport. It goes from 0 to 100 kilometers per hour (or 27.8 meters per second) in 2.2 seconds. This translates into an acceleration of:

a = 27.8 meters per second / 2.2 seconds = 12.6 m/s²

This is more than the free fall acceleration! To be more specific, it’s 12.6 / 9.81 = 1.28 g. If you got $ 4,000,000 to spare, how about getting one of these? But even this is nothing compared to what astronauts have to endure during launch. Here you can see a typical acceleration profile of a Space Shuttle launch:

(Taken from

Right before the main engine shutoff the acceleration peaks at close to 30 m/s² or 3 g. That’s certainly not for everyone. How much can a person endure by the way? According to “Aerospace Medicine” accelerations of around 5 g and higher can result in death if sustained for more than a few seconds. Very short acceleration bursts can be survivable up to about 50 g, which is a value that can be reached and exceeded in a car crash.

One more thing to keep in mind about acceleration: it is always a result of a force. If a force F (measured in Newtons = N) acts on a body, it responds by accelerating. The stronger the force is, the higher the resulting acceleration. This is just Newton’s Second Law:

a = F / m

So a force of F = 210 N on a body of m = 70 kg leads to an acceleration of a = 210 N / 70 kg = 3 m/s². The same force however on a m = 140 kg mass only leads to the acceleration a = 210 N / 140 kg = 1.5 m/s². Hence, mass provides resistance to acceleration. You need more force to accelerate a massive body at the same rate as a light body.

For more interesting physics articles, check out my BEST OF.

Mathematics of Explosions

When a strong explosion takes place, a shock wave forms that propagates in a spherical manner away from the source of the explosion. The shock front separates the air mass that is heated and compressed due to the explosion from the undisturbed air. In the picture below you can see the shock sphere that resulted from the explosion of Trinity, the first atomic bomb ever detonated.

Great Formulas_html_m67b54715

Using the concept of similarity solutions, the physicists Taylor and Sedov derived a simple formula that describes how the radius r (in m) of such a shock sphere grows with time t (in s). To apply it, we need to know two additional quantities: the energy of the explosion E (in J) and the density of the surrounding air D (in kg/m3). Here’s the formula:

r = 0.93 · (E / D)0.2 · t0.4

Let’s apply this formula for the Trinity blast.


In the explosion of the Trinity the amount of energy that was released was about 20 kilotons of TNT or:

E = 84 TJ = 84,000,000,000,000 J

Just to put that into perspective: in 2007 all of the households in Canada combined used about 1.4 TJ in energy. If you were able to convert the energy released in the Trinity explosion one-to-one into useable energy, you could power Canada for 60 years.

But back to the formula. The density of air at sea-level and lower heights is about D = 1.25 kg/m3. So the radius of the sphere approximately followed this law:

r = 542 · t0.4

After one second (t = 1), the shock front traveled 542 m. So the initial velocity was 542 m/s ≈ 1950 km/h ≈ 1210 mph. After ten seconds (t = 10), the shock front already covered a distance of about 1360 m ≈ 0.85 miles.

How long did it take the shock front to reach people two miles from the detonation? Two miles are approximately 3200 m. So we can set up this equation:

3200 = 542 · t0.4

We divide by 542:

5.90 t0.4

Then take both sides to the power of 2.5:

t 85 s ≈ 1 and 1/2 minutes


Let’s look at how the different parameters in the formula impact the radius of the shock sphere:

  • If you increase the time sixfold, the radius of the sphere doubles. So if it reached 0.85 miles after ten seconds, it will have reached 1.7 miles after 60 seconds. Note that this means that the speed of the shock front continuously decreases.

For the other two parameters, it will be more informative to look at the initial speed v (in m/s) rather the radius of the sphere at a certain time. As you noticed in the example, we get the initial speed by setting t = 1, leading to this formula:

v = 0.93 · (E / D)0.2

  • If you increase the energy of the detonation 35-fold, the initial speed of the shock front doubles. So for an atomic blast of 20 kt · 35 = 700 kt, the initial speed would be approximately 542 m /s · 2 = 1084 m/s.

  • The density behaves in the exact opposite way. If you increase it 35-fold, the initial speed halves. So if the test were conducted at an altitude of about 20 miles (where the density is only one thirty-fifth of its value on the ground), the shock wave would propagate at 1084 m/s

Another field in which the Taylor-Sedov formula is commonly applied is astrophysics, where it is used to model Supernova explosions. Since the energy released in such explosions dwarfs all atomic blasts and the surrounding density in space is very low, the initial expansion rate is extremely high.

This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: You can take another quick look at the physics of shock waves here: Mach Cone.

A tunnel through earth and a surprising result …

Recently I found an interesting problem: A straight tunnel is being drilled through the earth (see picture; tunnel is drawn with two lines) and rails are installed in the tunnel. A train travels, only driven by gravitation and frictionless, along the rails. How long does it take the train to travel through this earth tunnel of length l?

The calculation, shows a surprising result. The travel time is independent of the length l; the time it takes the train to travel through a 1 Km tunnel is the same as through a 5000 Km tunnel, about 2500 seconds or 42 minutes! Why is that?

Imagine a model train on rails. If you put the rails on flat ground, the train won’t move. The gravitational force is pulling on the train, but not in the direction of travel. If you incline the rails slighty, the train starts to move slowly, if you incline the rails strongly, it rapidly picks up speed.

Now lets imagine a tunnel through the earth! A 1 Km tunnel will only have a slight inclination and the train would accelerate slowly. It would be a pleasant trip for the entire family. But a 5000 Km train would go steeply into the ground, the train would accelerate with an amazing rate. It would be a hell of a ride! This explains how we always get the same travel time: the 1 Km tunnel is short and the velocity would remain low, the 5000 Km is long, but the velocity would become enormous.

Here is how the hell ride through the 5000 Km tunnel looks in detail:

The red, monotonous increasing curve, shows distance traveled (in Km) versus time (in seconds), the blue curve shows velocity (in Km/s) versus time. In the center of the tunnel the train reaches the maximum velocity of about 3 Km/s, which corresponds to an incredible 6700 mi/h!