distance

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.

Advertisements

Length – From Electrons to Galaxies

Even compared to a hydrogen atom (with its diameter of 1 A = 1 angstrom), the electron is microscopic. It is about 0.00006 A in diameter, or in other words: you would need 16,700 electrons to fill the length of a hydrogen atom. A water molecule, which consists of two hydrogen atoms and one oxygen atom, measures roughly 3 A. The largest naturally occurring atom, uranium, goes up to 4 A and a common glucose molecule is around 9 A.

This is where we leave the realm of atoms and molecules, that can only be penetrated by hi-tech electron microscopes. With these, resolutions below 1 A can be achieved, making images of individual atoms possible. The good old light microscope can go as low as 2000 A and no further. But this is enough to observe individual bacteria (10,000 A) and human cells (100,000 A). The latter is already about one-tenth the width of a human hair (1,000,000 A = 0.0001 m). This means that now we are nearing the length scales we are familiar with.

The thickness of a credit card is around 0.0008 m, the average red ant is about 0.005 m long and one inch measures 0.025 m. From the length of a cigarette (0.1 m) over the height of a person (1.7 m) and the wingspan of the Boeing 747 (64 m), we quickly approach the high end of the length scale.

The tallest man-made structure is Burj Khalifa, a 830 m tall skyscraper in Dubai. While already mind-boggling, it dwarfs in comparison to the highest mountain on Earth, the mighty Mount Everest, with its height of 8848 m. From this height, it would take about two minutes to reach sea level in free fall. From the International Space Station (400,000 m) however, the Mount Everest is just a small bump in an enormous sphere of diameter 12,700,000 m.

Going into space, the distances quickly grow beyond our comprehension. The Apollo astronauts had to travel 380,000,000 m = 1.3 light-seconds to get to the Moon. Any mission to Mars has to travel one-hundred eighty times that (4 light-minutes). Multiply that by another factor of ninety, and you get to the former planet Pluto (350 light-minutes).

This is where things get crazy. To reach the next star, Alpha Centauri, you’d have to travel 4.2 light-years or about 550,000 times the distance Earth-Mars. The center of our home galaxy is roughly 10,000 light-years away, the nearest galaxy Canis Major Dwarf adds another factor four to that (42,000 ly). In the grand scheme of things though, even this is not that much. The light we we observe coming from the nearest spiral galaxy, Andromeda, has been traveling for a mind-blowing 2.5 million years.

Where does it all end? Nobody knows for sure. The farthest galaxy is z8_GND_5296, discovered 2013 by the Hubble telescope and Keck Observatory in Hawaii. It is 13.1 billion light-years away. This means the light we see has been sent into space long before Earth came to be. Maybe the galaxy does not even exist anymore, maybe all the stars within it are dead by now. We’ll have to wait another 13.1 billion years to see if that’s the case. I’ll update the post then.

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 http://www.russellwestbrook.com)

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.