## Wednesday, January 25, 2017

### Concepts of Acceleration

Before starting an exploration of acceleration, it is useful to revisit how we think about velocity. Previously, velocity was described like this:

average velocity: the value represented by the displacement of an object in relationship to the elapsed time. Remember that displacement is the measure of the length of a line lying directly between the initial position and a final position of an object that moved.

What is key in this is that velocity is the rate at which the position of an object changes. The equation that relates displacement to velocity and time is:

d = vt

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

To complicate things (because that's what we do) displacement can be re-imaged as a change of position—the difference between initial position and final position—and elapsed time can be re-imaged as the difference between initial time and final time.

Okay, so if complicated formulas intimidate you, it's okay to close your eyes for a few seconds!

The re-imaged equation becomes, where:

di = initial position (or beginning displacement from some reference point)
df = final position (or final displacement from some reference point)
ti = initial time (clock reading at the beginning, when the object is at di)
tf = final time (clock reading at the end, when the object is at df)
v = average velocity

(d - di) = v ( tf - ti)

Okay, you can open your eyes now!

The "ah ha moment" above is that velocity relates to how the position of the object changes. Using words instead of math symbols, the idea is a lot more logical.

Velocity is the rate that position changes over some period of time.

This could be stated another way with displacement being the subject. Think about it this way:

The change in position of an object is determined by its rate of motion (velocity) and how long it moves.

Further, if it is some distance from a reference point to begin with (di), it will end up that distance plus how far it moved. That is evident in this equation:

df = di + vt

(back to that, yeah…)

Bam!

On To Acceleration!

Now, with the understanding of velocity refined, understanding acceleration becomes much simpler. There is a perfect parallel between acceleration concepts and velocity concepts!

To begin, imagine sitting in a car at a red light. When the light turns green, the driver presses the gas pedal. WHICH is CALLED the accelerator because… Pressing the accelerator, the flow of fuel to the engine increases, the RPMs increase. The passengers of the car feel themselves pressed back into the seat as the car moves forward. The speedometer begins to show a change in speed

5 MPH
10 MPH
20 MPH
40 MPH

That image is more than is needed to understand the concept of acceleration!

Acceleration is the rate that velocity changes over some period of time.

The parallels between acceleration and velocity carry through to the math equations as well. Just to prove the point, the previous discussion of velocity is going to be copied below this sentence and all the relevant word changes will be made in red!

What is key in this is that acceleration is the rate at which the velocity of an object changes. The equation that relates velocity to acceleration and time is:

vf = vi + at

where v is change in velocity, a is rate of acceleration, and t is elapsed time.

To complicate things (because that's what we do) change in velocity can be re-imaged as the difference between initial velocity and final velocity, and elapsed time can be re-imaged as the difference between initial time and final time.

Okay, so if complicated formulas intimidate you, it's okay to close your eyes for a few seconds!

The re-imaged equation becomes, where:

vi = initial velocity (or beginning velocity from some reference point)
vf = final velocity (or final velocity from some reference point)
ti = initial time (clock reading at the beginning, when the object is at di)
tf = final time (clock reading at the end, when the object is at df)
a = acceleration rate

(v - vi) = a ( tf - ti)

Okay, you can open your eyes now!

The "ah ha moment" above is that acceleration relates to how the velocity of the object changes. Using words instead of math symbols, the idea is a lot more logical.

Acceleration is the rate that velocity changes over some period of time.

This could be stated another way with change in velocity being the subject. Think about it this way:

The change in velocity of an object is determined by its rate of acceleration and how long it moves. Or, to save pixels:

v = at

(back to that, yeah…)

So you can see that acceleration and velocity are similar, but that velocity relates to the change of position and acceleration relates to the change of velocity.

Another parallel concept between acceleration and velocity is when the object has an initial velocity and then undergoes acceleration. This is just like, with displacement, when an object is some distance from a starting point and begins to move. For velocity, the equation is:

df = di + vt

where df is the final distance from a reference point, di is the initial distance from a reference point, v is average velocity and t is elapsed time.

The relationship between final velocity and acceleration is found using a parallel equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is acceleration rate and t is elapsed time.

In many cases, the "final velocity" concept is easier to visualize than is "final distance." Look at the two examples that follow.

Final Distance
A car starts out 4 miles from the state line and travels at an average velocity of 20 miles per hour for 2 hours. How far from the state line does it end up?

For the above, the equation is (how far does it end up?):

df = di + vt

df = 4 mi + 20mi/hr • 2 hr

df = 4 mi + 40mi
df = 44 mi

While this is not terribly complicated, processing the idea of "how far from the state line" is a bit unfamiliar. The idea that the car moves 20 MPH for 2 hours is fairly simple—40 miles is traveled. So, the car moves 40 miles from the initial position, which is 4 miles from the state line, so… Well, in the end, it seems a little less obvious.

HOWEVER, with acceleration, the idea of final velocity is more readily accessible. The basis of the idea is that something is moving at a given velocity ( ) and either speeds up or slows down. Thus, it has a new, different velocity. Hence:

vf = vi + at

Final Velocity
A car starts out traveling at 40 mi/hr. It speeds up at a constant rate of 10 mph/sec for 2 second. How fast does it end up traveling?

For the above, the equation is (How fast does it end up traveling?):

vf = vi + at
vf = 40 mi/hr + 10 (mi/hr)/sec • 2 sec

vf = 40 mi/hr + 20 mi/hr

So the change in speed is + 20 mi/hr. Thus, the final velocity is…

vf = 60 mi/hr

In many cases, the initial velocity will be zero. This will be true whenever the situation includes an object "at rest" or "not moving." Here are some phrases to look for that allow you to assume the initial velocity is zero:

An object is at rest…Two stationary zebras…The airplane begins to move…

Generally, unless there is some definite indication of an initial velocity, you can assume vi = 0 and if so, the equation reduces nicely to:

vf = at

However, it is good practice to start with the complete equation each time, fill in the give values, and then solve. One more step is not going to kill you!

## Tuesday, January 24, 2017

### Elapsed Time

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.

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"

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.

## Monday, January 16, 2017

### Motion with Constant Velocity or Speed

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.)

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.

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?

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.

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
d= df - vt
t = (df - di)/v
v = (df - di)/t

_________________________________

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.

_________________________________

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 in opposite directions. 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.