Finding Time Using the Distance Equation

Any calculation of time (t) is relatively easy in the cases where either vi or a was zero. In the cases where neither are zero, the math results in a second degree polynomial equation such as:

0 = x2 + 3x - 12

Where this occurs in physics of motion is in the full distance equation:

df = di + (vi)(t) + (1/2)(a)(t2)

Although this does not exactly match the expected form for a quadratic equation, it can easily be rearranged as such:

df = di + (vi)(t) + (1/2)(a)(t2)

0 = di - df + (vi)(t) + (1/2)(a)(t2)

0 = (1/2)(a)(t2) + (vi)(t) + di - df

Keep in mind that di - df will yield a number when the values are plugged in and simplified.

To solve these problems, the steps are the same for any problem in science, BUT the algebra becomes harder.

After you have plugged in the numbers, combine like terms and simplify. Suppose the following:

df = 89

di = 50

vi = 4

a = 6

and you need to find t

df = di + (vi)(t) + (1/2)(a)(t2)

89 = 50 + (4)(t) + 1/2(6)(t2)

0 = -39 + 4t + 3t2

(Put it in normal quadratic form.)

0 = 3t2 + 4t - 39

Now you can either factor or use the quadratic formula to find the values for t:

0 = ( t  - 3 )( 3t + 13)

0 = t - 3    AND   0 = 3t + 13

3 = t        AND   -13/3 = t

Since time cannot be negative within the context of classical physics, only t = 3 is a valid answer.

Using the quadratic equation will yield the same results.

While it is far easier to find t when either a or vi is zero, the math to find t when that is not the case is not beyond the skills of a student taking an introductory physics class.

For another look at this process, check out this video:

https://youtu.be/_G7zAX98Tw0

​

Physical Quantities: Quantities and Units

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Where are we going with this? This page will assist in understanding the relationship between units and the quantities they measure.

When looking at the world around, people pretty much automatically attend to physical quantities, pretty much without thinking about them. The youngest child intuitively can judge differences with some skill.

By the time the child reaches school age, the following conversation would seem drastically out of place:

Teacher: "Mary, how far can you run?"
Mary: "I can run six pounds! Very heavy!"

Almost all children would know that far and heavy describe different physical quantities.

As most readers of this would know, one of the jobs of science is to complicate things. No… not that… 

to clarify things by assigning specific words with specific meanings to specific ideas, objects and concepts in order to make discussion precise and accurate (which are both words surrounded by confusion, ironically).

Therefore, with regard to physical quantities, a certain set of words are used to describe specific things.

Included in those sets of words are two sides to every quantity. 

One side is the concept of what is being described. Mary, in the dialog above, gave an answer about one quantity with words used for a different. 

The other side of the concept is, for every quantity, specific ways that it is measured (which are called units). Pounds do not go with a "how far" question!

First what are the things being described?

Types of Physical Quantities

The following list is certainly not all-inclusive. It includes a few very common and familiar quantities that are measured.

Distance: Distance is a measure of… okay, distance is so common it actually has different meanings, so in science, there are different words that are used so in order to be more specific.

Distance (Most General): The measure of how far two points are from each other, as in, "The tip of the antenna was 12 meters from the the surface of the window."

Distance (A specific case) Think Distance Traveled: The length of a path that a moving body takes, as in, "The dog ran from tree to tree through the park until it was at the tree next to the one where it started, covering a distance of 250 yards." Compare with Displacement, next!

Displacement: The measure in a straight line between where a moving body begins and ends, as in, "The dog ran from tree to tree through the park until it was at the tree next to the one where it started, resulting in a displacement of only 4 yards."

Length, Width, Height: Pretty much what you expect, on this! The measure of how far specific points of an object are from other points on the object.

Other things that would be included in the distance concept are circumference, radius, perimeter, range, altitude, depth… You can probably think of others. But will you?

In general, distance is the measure of how far apart two things are.


Volume: (Not talking about sound, which borrowed from this concept for its own purposes!) This also is pretty much what you expect!

Volume is the space in three dimensions that an object or substance takes up (or holds). If a tank holds 25 gallons of gasoline, then the volume of the gasoline that filled the tank would be (duh) 25 gallons. A bottle of soda holds 20 ounces. A different bottle holds 2 liters.

Mass: Mass is a physical quantity that is the result of how many protons, electrons, and neutrons are all in a given space. It is actually not directly observable, though there are devices (scales and balances) that combine with gravity (or other acceleration) and other laws of physics so that it can be measured. Mass is a measure of the amount of matter an object contains.

If two things with the same volume have different masses, one would feel heavier than the other.

Good news: There are scales that measure mass, so you can just plop things down and get a number!

Weight: Weight is a measure of mass under a particular condition. Many people have heard things like, "Well, on the moon, I'd only weigh 12 pounds."

Weight is a basic concept that people are very familiar with. Weight is the degree of heaviness something has.

Time: This is something people very intuitively understand, but which is actually a very abstract concept. Stop reading and write down a definition of time. Not how time is measured! Try a definition of time that does not use seconds, minutes, hours, days, etc. and see what you come up with!

According to the Oxford dictionary, time is the indefinite continued progress of existence and events in the past, present, and future regarded as a whole. 

Time is a very basic concept in science, and fortunately, our intuitive understanding is enough for us to use it.

How do you measure these quantities?

The list below will connect the quantities above with the SOME OF THE units used to measure them. Common units in science are in bold and "normal" abbreviations for the common units are in parenthesis..

NOTE: Italics units are from the English system

Distance: inches, feet, yards, miles, meters (m), kilometers (km), centimeters (cm), light-years

Volume: gallon, ounce, cup, teaspoon, liters (l or lt), milliliters (ml)

Mass: slugs, grams (g or gr), kilograms (kg)

Time: hours (hr), minutes (min), seconds (s)

A few more…

Some units are derived or combined from the basic units but are so common they are worth noting here.

Force: Newtons (n or N), Pounds

Weight (which is a force) is also measured in pounds, newtons

Speed or Velocity: MPH, m/s, cm/s, km/hr
Speed or Velocity is a distance unit divided by a time unit.

Acceleration: (m/s)/s, MPH/s, (km/s)/s, cm/s²  m/s²

Acceleration is a velocity unit divided by a time unit.

At a glance…

System	Mass	Weight	Distance	Time	Acceleration Due to Gravity
English	Slug	Pound lb	Feet ft	Seconds s	32 f/s/s
Metric	Kilogram kg	Newton N	Meters m	Seconds s	9.8 m/s/s

And also…

Temperature: Temperature is, within the kinetic theory of matter model, defined as the average kinetic energy of the molecules in a system or substance. How much thermal energy is present? How hot is it? The absence of thermal energy is described as being cold. Temperature is measured in degrees. In science, officially in Kelvin degrees (°K), but often in Celsius degrees (°C). 

SI Units

With so many units, things could get confusing. So… Some smart people made a decision that, a a general best-practice science would be built on seven base units. They also gave it a fancy name:

The International System of Units

Then… they decided they would abbreviate it… from French 

Système international (d'unités)

The seven base SI units and some corresponding constants allow meaningful and uniform dialog about all matters of science. 

https://en.wikipedia.org/wiki/International_System_of_Units

Displacement and Acceleration

If the relationship between displacement, acceleration, and elapsed time is extended to the most general case, the following equations need to be combined.

1.)

df = di + vt

where df is the final displacement from the fixed point, di is the initial displacement from the fixed point, v is rate of change in position, (average velocity), and t is elapsed time.

2.)

Vf = Vi + at

where Vf is final velocity, Vi is initial velocity, a is the rate of acceleration, and t is the elapsed time.

3.)

Vave = (Vf  + Vi) / 2

where Vf is final velocity, Vi is initial velocity, and Vave is the average velocity.

To begin, start with the general displacement equation:

df = di + vt

Understanding that v is average velocity, the equation becomes:

df = di + ((Vf  + Vi) / 2) • t

And since Vf can be found in relationship to acceleration, the following emerges:

df = di + (((Vi + at) + Vi) / 2) • t

Solving the problem in steps generally is more easily understood. To find displacement when acceleration is present, do the following.

Step 1.) Find the final velocity: Vf = Vi + at

Step 2.) Find the average velocity: Vave = (Vf  + Vi) / 2

Step 3.) Find the displacement: df = di + vt

EXAMPLE

A model car is 3 meters from the starting line on a model car race track and it is moving at 2 m/s. It accelerates at a rate of 5 (m/s)/s for 4 seconds. How far does it end up from the starting line?

Step 0.) Collect the data!

di = 3 m

vi = 2 m/s

a = 5 (m/s)/s

t = 4

Step 1.) Find final velocity.

Vf = vi + at

vf = 2 + 5•4

vf = 2 + 20

vf = 22

Step 2.) Find average velocity.

v(ave) = (vf + vi)/2

v(ave) = (22 + 2)/2

v(ave) = 24/2

v(ave) = 12 m/s

Step 3.) Find the total displacement.

df = di + vt

(remember v is average velocity and t is the same elapsed time as for the acceleration)

df = 3 + 12•4

df = 3 + 48

df = 51 meters 

So, the model car ends up 51 meters from the starting line.

Elapsed Time

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Elapsed Time

How long did it take? How much time passed?

The answer to these types of questions is answered by finding the elapsed time.

Let's start with a definition:

elapsed time: the difference between the final time and the initial time.

To find out how much time passes, we can subtract the clock or stopwatch reading at the beginning of the event from the clock or stopwatch reading at the end. With that in mind, 

initial time is the clock or stopwatch reading at the beginning of an event. For a stopwatch, it could often be zero.

final time is the clock or stopwatch reading at the end of an event.

For example, a track coach starts a stopwatch when the gun fires and stops it when the runner crosses the finish line. The initial time reading was zero and the final reading on the stopwatch was 10.5 seconds. Thus, the time that the runner ran was 10.5 seconds.

Elapsed time will always be expressed with time units, such as seconds, minutes, hours, etc.

Calculating elapsed time is generally easy to understand. If your movie starts at 7:00 and ends at 9:15, how long does it last? Or, in other words:

What is the elapsed time of a movie that begins at 7:00 and ends at 9:15?

Most people will almost instantly say 2 hours and 15 minutes. But wait! There is actually a math way to do this with math!

If elapsed time is t and initial time is ti and final time is tf, then:

t = tf - ti

Of course, having a formula spoils the simplicity of the concept, but it also allows for the concept to be applied in non-intuitive situations. But, for now, the example above:

t = what we are looking for

tf = 9:15 o'clock

ti = 7:00 o'clock

   9:15

- 7:00

______

  2:15 hours

The process becomes more complex when you have to "borrow" because, time. A minute is 60 seconds. An hour is 60 minutes. So when you borrow, you have to borrow 60 of the other thing.

Thus, the problem goes like this:

t = what we are looking for

tf = 9:15 o'clock

ti = 7:30 o'clock

   9:15

- 7:30

______

????

When you borrow 60 minutes so that you can subtract, you get this:

   8:75

- 7:30

______

  1:45 = 1 hour and 45 minutes.


For examples worked out and more…

	Scan the QR code to open movie.
Finding Elapsed "Clock Time"	Scan the QR code to open movie.

This can be extended to days, as well, though with less clarity. (What do you mean by "day"? Is it 24 hours or is it the actual amount of time that it takes for the earth to rotate once, which is more than 24 hours?)

When reading a stopwatch, you will see the readings written like this:

HH:MM:SS.fs

Where HH is hours, MM is minutes, SS is seconds and fs is fractions of a second.

The math works the same—if you borrow, you have to borrow in 60s.

Motion with Constant Velocity or Speed

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Motion with Constant Velocity or Speed

Motion with constant velocity or speed is, mathematically, a basic rate problem. That is to say that something occurs  regularly over a period of time, and there is an outcome that is in relationship with a rate of "going on" and how long something goes on.

For example, if a cookie machine can turn out 10 cookies every 15 minutes, then the total number of cookies will be determined by how long the cookie machine runs. The rate can be stated as 40 cookies per hour, so after 3 hours, there will be 120 cookies.

Motion with a constant velocity or speed works the same way. The rate will be how far something moves in a given time—which is called speed or velocity—and time will be how long it moves.

In most cases, calculating any part of the relationship is relatively simple. It is not that hard to figure out a formulaic relationship by just looking at the units. (See Video.)

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A speedometer shows speed to be some number of miles per hour or kilometers per hour. So:

speed = KPH

The "per" means divide, so…

speed = k/h

where k is kilometers and h is hours. Changing to the variables normally used, the distance equation is:

d = vt

where:

d is displacement, v is average velocity, and t is elapsed time.

Similarly, the same equation works for motion NOT along a straight line but with these changes:

d is distance traveled, v is average speed, and t is still elapsed time.

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EXAMPLE 1:

A car travels at an average speed of 35 miles per hour for two hours. How far does it travel?

Begin by identifying the needed values:

d = what is being looked for 

v = 35 miles per hour 

t = 2 hours

d = vt 

d = 35 m/hr * 2 hr

d = 70 miles (the 'hr's cancel out)

EXAMPLE 2:

A car travels at between two points that are 10 miles apart in a straight line. The trip begins at 8:00 and ends at 10:00. What is average velocity?

Begin by identifying the needed values:

d = 10 miles 

v = what is being looked for

t = final time - initial time

t = 10:00 - 8:00

t = 2 hours

d = vt 

10 miles = v * 2 hr

10 miles/2 hours = 

5 m/hr = v

The math for motion with constant velocity or speed is very easy, but rearranging the formula for each case could benefit some people. Thus, each part of the formula can be found using these three equations:

d = vt

v = d/t

t = d/v

Based on what is given in the problem, the math involves only multiplying or dividing, in most cases.

Naturally, there has to be a more complicated version of motion with constant velocity or speed. Nevertheless, it is not all that much more complex. It just looks more complex!

The more complex version asks how far something is from somewhere if it starts some distance away from some other place. IKR?

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Here is what that looks like:

A canoe left the shore 10 miles downstream from a bridge. It traveled at 2 miles per hour for 3 hours. How far was it from the bridge?

Now, to reason this out with logic alone, it is fairly easy. The canoe travels at 2 miles per hour for 3 hours, so it goes 6 miles. However, it STARTED 10 miles from the bridge, so it ends up a TOTAL of 16 miles from the bridge. BAM!

But the "official physics" formula uses subscripts. So… there's that. Or this:

df = di + vt

where:

df is final distance/displacement (this is sometimes dt for total distance)

di is initial distance/displacement

v is average speed/velocity

t is time

Back to the canoe problem above, we have:

df = what we want to find

di = 10 miles from the bridge = 10 miles

v = 2 miles per hour

t = 3 hours

df = di + vt

df = 10 miles + 2 m/hr * 3 hr

df = 10 miles + 6 miles

df = 16 miles

The subscripts just make it look harder, but it is really just the same logic and reason. The math is just as easy—addition and multiplication.

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It is not really hard to rearrange the more complicated formula to solve for any of the four values. This is what it would look like:

df = di + vt

di = df - vt

t = (df - di)/v

v = (df - di)/t

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While there are, indeed, formulas that model motion with constant speed or velocity, it is most important to remember that ration and logic always apply. It is tempting, at times, to give up because the formula is intimidating, but there is no need to do that. Remember that the formulas are just a way to get to the same conclusions that can be found by simply reasoning it out.

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When TWO Things Are Moving

It is very common to be interested in what's happening where more than one thing is moving. When two things are moving, there are a variety of ways to solve problems.

A very easy way to work such problems is to construct a frame of reference that reduces the complexity greatly.

If the two objects are moving on the same axis… trains on parallel tracks, vehicles on the same road… it is often possible to look, not at the individual velocities or speeds, but rather to look at closing speed.

Closing speed is the speed (or velocity) that would exist if the frame of reference was created around one of the moving objects. It is relatively intuitive.

• A car and a truck are 500 feet apart traveling in the same direction. The car is behind the truck traveling at 100 feet per second and the truck is traveling at 50 feet per second. How long does it take for the car to close the gap between the two vehicles?

This is easily solved by looking at the difference in speeds. The 500 foot gap will be covered at a rate of 50 feet per second, the difference in the speed of the car and truck.

• A car and a truck are 500 feet apart traveling toward each other. The car is traveling at 100 feet per second and the truck is traveling at 50 feet per second. How long does it take for the car to close the gap between the two vehicles?

This is easily solved by looking at the difference in speeds. The 500 foot gap will be covered at a rate of 150 feet per second, the sum of the speed of the car and truck.