In physics, we are concerned both with objects at rest (static) and objects in motion (dynamic). Speed is an important measure of the quantity and character of motion. In math classes, we often refer to speed as rate, or rate of travel. Speed is defined as the distance traveled divided by the time it takes
It's worth pausing here to note that physics is full of three-variable equations like this, and much of the work you'll do to model and solve physical problems will consist of rearranging and solving equations like this with simple algebra. Here are the three rearrangements of the speed equation and how they're obtained:
There's a useful trick to these kinds of rearrangements if your algebra is weak or rusty, the "magic triangle" (below). You can use it to find the rearrangements of any three-variable equation. Just set up the equation in the triangle in a form you know - note the similarity of the arrangement in the left-most triangle: speed equals distance over time. Then to solve for one of the variables, cover that variable with a finger and read off the solution. The middle triangle: distance = speed times time. The right triangle: time = distance over speed.
If you drive from Salida, CO to Denver, CO, you'll cover a distance of about 150 miles (240 Km), and it will take you about 3 hours. What you can calculate from that is average speed:
[Note: in science, we don't really use the common abbreviation "mph" for "miles per hour."]
It's important to notice that what we've actually calculated is an average speed. In fact, on that trip, there are several stop signs, mountain passes and other places where speed can be faster or slower than the average. We don't know anything about any of those details when we calculate average speed, just the total time and distance.
Instantaneous speed, the speed of an object, like a car, at any one instant in time is harder to calculate (it requires calculus – see the box below), but we can measure instantaneous speed at any time in a car by looking at the "speed-o-meter."
Often we put a bar over a quantity to denote that it is an average, so average speed is sometimes written as s.
Most of the time in physics we are interested in average speed.
Solution: First we need to convert 1 hour + 50 min. to hours. 50 minutes is 5/6 hour or 0.833 hours, so 1:50 = 1.833 hours.
Now the speed average speed is just:
We should probably round that to 15.9 or 16 mi./h because we only know the distance to the tenths place – it's just good practice.
Note: Once in a while you'll see mi./h written as mi.h-1. It's written that way because printed text like what you're reading now looks best on one line, and because it's bad practice to write fractions with a slash / on the same line – it's a good way to lose track of units.
Solution: If you are like most students, you'll want to start plugging in numbers right away, but I strongly urge you to resist that. Organize the equation to solve for the variable you're looking for first. The speed-distance-time equation is:
To solve for time, first multiply both sides by time. A good rule to follow is: If the variable for which you want to solve is in the denominator, first ... get it out of there.
Now divide both sides by the speed to get:
Now we can plug in distance and average speed to get the time. Pay attention to the units. They can tell you when you're messing up.
One of a number of commonly-used trans-America bike routes.
Notice that dividing by mi./h is the same as multiplying by the reciprocal, so the units of miles divide away nicely
The resulting time in hours is:
That's about 22 8-hour days.
Solution: First we need to calculate the average speed. It's just the numerical average of the two speeds that we know, 0 m/s and 6000 m/s:
Now we need to get the times on the same scale. Let's use seconds, so we'll have to convert years to seconds:
It's usually better to use scientific notation when numbers get so big. Now the definition of speed is:
We rearrange before plugging in numbers. Solving for distance gives us:
Now plugging in numbers and canceling units gives the distance.
That's a lot of Kilometers.
When distance traveled is plotted vs. elapsed time, the slope of the graph (rise over run) is the average speed.
Often in math we get the impression that units and even the relative scales of the y and x axes (distance and time here) must be the same. It isn't so. In this graph, distance in meters divided by time in seconds gives speed in m/s, an acceptable unit.
The graphs below illustrate a few different kinds of d vs. t graphs you might encounter. You should learn to interpret graphs like these.
An object moves at constant speed, stops and moves at the same speed in the opposite direction.
An object moves, stops, moves again in the same direction, stops and moves again.
An object moves in one direction with increasing speed (acceleration).
Often when we're working with speeds, we're working with relative speeds, or the speeds (or rates) of two things that are linked. There are some nifty techniques for dealing with problems of this kind.
Here are a couple of examples of such problems. They make good use of your organizational and algebra skills.
Solution: Our strategy will be to find the speed of each runner (because we have both distance and time), then use those speeds to calculate how long it will take runner A to run the full 10,000 m, then use that time to calculate how far runner B will go. The difference between that distance and 10,000 m will be our head start.
The speed of runner A is
and the speed of runner B is
I've converted these speeds to meters/second because it's a very common unit, but that's not absolutely necessary.
Now we can rearrange the speed equation to solve for time:
The time it takes for the faster runner to cover 10,000 meters is
That's just a little over 39 min. In that time, runner B will cover a shorter distance. Rearranging the speed equation again to d = st, we get:
So in order to finish in a tie, runner A will have to let runner B have a 10,000 - 8,750 = 1,250 m (or 1.25 Km) head start.
Solution: It doesn't seem like we have enough information to solve this problem, but we do. Two key things help us. First, running around the track, we have the distance each runner covers in the same time. that will allow us to link up equations through the common time using the transitive property. Second, in the 5,000 m race the faster runner (A) runs the distance that B runs plus some unknown distance.
We take advantage of the same time on the track run by calculating the speeds each runner:
Solving for time, t = d/s, we can eliminate the time to get:
Cross multiplication gives us
and solving for sB tells us that the speed of runner B is 11/13 the speed of runner A. We don't know the actual speeds, but we do know the relationship between them.
Now the graph of distance vs. time is a line, for which the slope is the speed. From this we can infer a couple of facts:
Finally, the difference is 2/13, so we have the head-start distance:
The transitive property of algebra says that if two things are equivalent to the same thing, then they are equivalent to each other.
This section is entirely optional, but read on if you know something about calculus and you'd like to learn about its role in physics. By the way, if you don't know anything about it, don't worry. Calculus really isn't a huge mystery, and it will make perfect sense when you do learn it.
Because the first derivative of a function is its slope, the first derivative of the position (x) with respect to time is the speed.
Now if we view the integral as a sum, then the net distance traveled, provided we have some functional form for the speed, is just the integral of that function between two times, t1 and t2
There are many other derivative and integral relationships in physics. Any rate is the derivative of some underlying function, and any time a quantity that can be represented as a continuous function is summed – that's an integral.
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