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 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.)
Scan the QR Code To Open Movie
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 normallyused, 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.
Scan the QR Code To Open Movie
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?
Scan the QR Code To Open Movie
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.
Scan the QR Code To Open Movie
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
_________________________________
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 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.
A large part of physics is the study of motion. Because words related to motion are very specific, it is worthwhile to examine them closely.
At the most basic level, motion involves something moving. The thing moving might be a person, a car, a horse, a plane, a train, an electron, a planet, a moon, a galaxy, a… it could be anything that moves, even an imaginary point such as "where the arrow points."
Often whatever is being examined or studied or looked at is generalized to be called an object. Instead of saying "an athlete runs…" it could be said "an object moves…"
object: (in the context of motion) the thing or things that is being observed.
Usage:
an object moves in a circle…
an object moves in a straight line at a rate of.
The location of an object is called its position. How an object changes position over time is generally the purpose of a study of motion.
position: the location of an object
With an understanding of object and position, it is possible to come up with a definition of motion:
motion: when an object changes position
If the idea of changing position is examined, three more specific ideas emerge. However, it is necessary to first look at another idea that will help make the ideas about changing position less complex.
The idea is that observing motion means that something starts and ends. As one example, you might want to think about how long it takes to go from home to school. You would look at the changes that occur between the time that you leave home until the time you get to school. This period of observation can be, for convenience, be called an event. Thus, motion could be discussed as what happens to an object's position during an event.
Returning to the concept of position, then, the understanding of the idea of events make the following definitions easier to phrase:
initial position: the location of the object at the beginning of an event.
…answers questions like "Where did it begin?"
position after…: the location of the object after some passage of time. For example, "position after 10 seconds" or "position after 4 hours."
final position: the location of the object at the end of an event.
…answers questions like "Where did it end up?"
Scan QR Code To Open Video
Consider this description of a motion event:
A train leaves the station at 7:00 a.m. and heads down the track at 50 miles per hour. How far would the train have traveled after 3 hours? The next station is 350 miles away. How long would it take to reach the next station.
Based on the description, there are three pieces of position information:
initial position : : "leaves the station" : : P1 or Pi
position after… : : "after 3 hours"
final position : : "reach the next station" : : P2 or Pf
Scan QR Code To Open Video
NOTES: Positions are places. Sometimes these places are described in relationship to other things, like 10 meters from a mark or 4 miles from a bridge. Positions are not distances. Distances (to be discussed next) have some measure and units, like 10 miles or 40 mm.
Distance and Displacement
The next concepts to consider are distance and displacement. Neither of these are particularly difficult to understand, but they are very specific and need to be considered very exactly.
distance or distance traveled or distance moved: the measure of the length of the path an object takes as it moves between an initial position and a final position
"Distance is the curvy line."
displacement: the measure of the length of a line lying directly between the initial position and a final position of an object that moved.
"Displacement is the straight line."…distance and displacement answers questions like "How far did it move?"
Scan QR Code To Open Video
Calculating the change in distance or the change in displacement is easy. These are the differences between initial and final position and is found as:
∆d = df - di
The same formula rearranged…
df = di + ∆d
where ∆d is the change in distance/displacement and df is the final distance and di is the initial distance.
In some cases (such as when position is given as a distance from some reference point) it makes more sense to calculate distance and displacement based on initial and final positions:
∆d = Pf - Pi
where ∆d is the change in distance/displacement and Pf is the final distance and Pi is the initial distance.
Measuring Time
Besides an object changing position, any study of motion requires some way to describe how long something takes. That is to say, we need to know how much time passes. The passing of time is commonly measured in hours, minutes, seconds, and fractions of seconds.
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.…answers questions like "When did it begin?"
final time is the clock or stopwatch reading at the end of an event.…answers questions like "When did it end?"
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 was zero and the final on the stopwatch was 10.5 seconds. Thus, the time that the runner ran was 10.5 seconds.
This allows for the understanding of a key concept related to motion: elapsed time.
elapsed time: the difference between the final time and the initial time.
In science time will usually be in hours, minutes, seconds, and fractions of a second.…answers questions like "How long did it take?"
Scan QR Code To Open Video
To find the elapsed time, subtract the initial time from the final time. This is not always easy due to the need to borrow across the colon, and is discussed in more detail elsewhere. However, the general case is described by: t = tf - ti where t is the elapsed time, tf is the final time and ti is the initial time.
NOTES
Initial time and final time are clock readings.
Elapsed time has a number of hours, minutes or seconds.
Measuring the Rate of Change in Position Finally, understanding the above concepts allows getting to the definitions of a few more important motion concepts:
speed: the instantaneous measure of the rate at which position is changing without regard to displacement or direction.
velocity: the instantaneous magnitude and direction of an object that is changing position
average speed: the value represented by the distance an object moves in relationship to the elapsed time.
average velocity: the value represented by the displacement of an object in relationship to the elapsed time.
…answers questions like "How fast was it going?" and "What was it's average velocity?"
Look at the following image:
The red line represents the actual path that an object took in traveling from A to B. If the red line is 50 feet long, then the distance traveled is 50 feet.
The blue line represents the displacement. It is the direct path between point A and B. If the blue line is 20 feet, then the displacement is 20 feet.
Assume that, for the motion from A to B, the elapsed time is 5 seconds.
The average speed would, therefore, be 10 feet/second.
The average velocity would, therefore, be 4 feet/second.
