We saw in the last section that a time-dependent position vector represents an object in motion. To define the motion in terms of speed and direction of motion, we need to take a derivative of the position with respect to time. This is called the velocity, or: [br][br][center][math]\vec{v}=\frac{d\vec{r}}{dt}.[/math][/center]The magnitude of the velocity is the [b]speed [/b]of the object, or how fast it's moving. Go back to the last section or the previous chapter if you don't recall how to calculate the magnitude of a vector.  Also, please do not confuse velocity and speed as is often done in ordinary language by non-technical people.  [br][br]To find the direction of motion when given a velocity vector, we have to do a little work. The idea is that we want to break a vector apart into its magnitude and direction, as we've seen before. The magnitude is a scalar quantity, and the direction is a vector quantity, and in this context needs to be a unit vector. We want to be able to break up the velocity like this:[math]\vec{v}=|\vec{v}|\hat{v}=v\hat{v}[/math]. The scalar term is the speed and the second term (the unit vector) is the direction of motion. Note that it is common to use shorthand to denote the magnitude, so [b]we are to understand that a vector without the vector arrow above it, means the magnitude[/b], or [math]|\vec{v}|=v[/math]. As before, to find the direction (unit vector), we simply divide the vector by its own magnitude, or [math]\hat{v}=\frac{\vec{v}}{|\vec{v}|}=\frac{\vec{v}}{v}[/math]. [b]The unit vector retains the same direction as the original vector but will be only one unit long[/b].

Since we will be using m/s as our unit of velocity (and speed), it would be nice to have a feel for how fast something like 10 m/s actually is. Before we move on, try converting a familiar unit of speed (in the US) to this less familiar unit of m/s. How many m/s is 60 mph (really mi/hr)? There are 1609 meters per mile.[br][br]You should have found that 60mph is 27 m/s. So roughly there is a factor of 2 between them, or even better 2.2 (still not exact, but quite close). Do you know of another metric unit that is 2.2 times its English counterpart? If you do, it will make memorizing this new, approximate conversion easier. Having a feel for the m/s unit should help you visualize problems you encounter in the future. The answer to the question above is that the other familiar conversion is for mass: 1kg=2.2lb.[br][br][color=#1e84cc]AN ASIDE: Can you multiply something by 2.2 in your head easily? Even if you can't, I bet you can multiply it by 2 rather easily. Of course you know that 10(2)=20. Well you also know that 10(0.2)=2 = 20/10. Notice then that to multiply something by 2.2 just means to multiply it by two, and then to add one tenth of that result to the first result. Here are some examples: 30(2.2)=60+6=66, or 45(2.2)=90+9=99, 55(2.2)=110+11=121, etc[/color]. [br][br]This is a quick way to convert kilograms to pounds in your head, or to convert m/s to mi/hr (approximately) in your head. A more precise conversion from m/s to mi/hr is 2.237, but our factor of 2.2 is less than 2% error. Fine for most purposes!

To get the velocity of an object from its position vector is as simple as taking a derivative. For instance, for an object with a constant position vector of [math]\vec{r}=2\hat{i}+3\hat{j}[/math], the velocity is [math]\vec{v}=0\hat{i}+0\hat{j}[/math] because the derivative of both 2 and 3 (constants) with respect to time is zero. If you're wondering about the product rule, since the first term is really [math]2\hat{i}[/math], and not just 2, you are correct to wonder. In this case the unit vector [math]\hat{i}[/math] does not change direction in time, nor will its magnitude change since it's a unit vector. So the derivative of the unit vector with respect to time is zero. This will not always be true of unit vectors. Sometimes they change direction in time and we will have to carefully evaluate the derivative. We will see such cases when we discuss circular motion in more detail. [br][br]Given a position vector that is time-dependent, such as [math]\vec{r}=4t\hat{i}+3\hat{j}[/math], we obtain a non-zero velocity vector. We get [math]\vec{v}=4\hat{i}+0\hat{j}[/math]. The x-component comes from [math]\tfrac{d}{dt}4t=4[/math] and the y-component is zero since the derivative of a constant always is. This tells us that the object is moving at 4m/s in the positive x-direction. Notice that the velocity is constant and not time-dependent. This naturally is the case since the position vector depends linearly on time, and the derivative (think slope) of a linear function is constant. Notice that the units of velocity are m/s, or how many meters of distance you are traveling per second of time elapsed.

If you think that we should be able to get position by simply integrating the velocity, you are mistaken. This follows directly from calculus. Let's start with the definition of velocity and see what happens when we integrate with respect to time: [br][br][center][math]\vec{v}=\frac{d\vec{r}}{dt} \cr[br]\vec{v}dt = d\vec{r} \cr[br]\int_{t_1}^{t_2} \vec{v}dt = \int d\vec{r} = \vec{r}_2-\vec{r}_1 = \Delta\vec{r}\cr[br][/math][/center]So we see that [b]integrating the velocity with respect to time gives us the displacement and not the position[/b]. This is sensible since an integral is just a sum. A sum of little changes [math]\int d\vec{r}[/math] should naturally give the total change of the vector ([math]\Delta\vec{r}[/math]), not the value of the vector ([math]\vec{r}[/math]). To find the later position of an object in this case, we'd need to know the earlier position and add it to the displacement, or [br][br][center][math]\vec{r}_2=\vec{r}_1+\Delta\vec{r} [br][/math].[/center]Notice that the earlier position, [math]\vec{r_1}[/math], is a separate piece of information that cannot be obtained from the velocity function!  On the other hand, by definition, the velocity can always be obtained from the position function with no additional information by simply taking a derivative. This is noteworthy. What that means is that if you take any function, solve for its derivative, and then integrate that result, you do not in general arrive back at the original function. Information is lost in taking the derivative! The derivative only tells you about the rate of change, but nothing at all about the value of the function. In the same way, there is more information in a time-dependent position function than there is in a velocity. The position function contains information about where something is, and with a derivative we can also use the position to find velocity. There is no recourse for finding position from velocity. The information is not there. You can find the CHANGE IN POSITION, or displacement, but not the absolute position. The missing value of the initial position vector must be provided separately.

[i]EXAMPLE: Position from Velocity[br]Consider a ping pong ball that starts at position [math]\vec{r}=1.5\hat{j}m[/math] and starting at [math]t=0[/math] has a velocity of [math]\vec{v}=1.0m/s\hat{i}+(5.0m/s-10m/s^2t)\hat{j}[/math]. Where will the object be at [math]t=0.5s[/math]?[/i][br][br][i]SOLUTION: We just need to integrate the velocity over the time interval to find displacement and then add the result to the initial position that is given. Integrating a vector quantity is just done term by term, just as differentiating a vector quantity is: [br][br][center][math]\Delta\vec{r}=\int_{t=0.0s}^{t=0.5s}[1.0m/s \hat{i}+(5.0m/s-10m/s^2t)\hat{j}] \; dt= \\[br][1.0tm/s \hat{i}+(5.0tm/s-\tfrac{10t^2}{2}m/s^2)\hat{j}]|_{t=0.0s}^{t=0.5s} \\[br]\Delta\vec{r}=0.5m\hat{i}+1.25m\hat{j}\\[br]\vec{r}_f=\vec{r}_i+\Delta\vec{r} \\[br]\vec{r}_f=1.5m\hat{j}+0.5m\hat{i}+1.25m\hat{j}\\[br]\vec{r}(t=0.5s)=(0.5\hat{i}+2.75\hat{j})m. [/math].[/center] [br][/i]

Recall from your studies of calculus that the average value of a function over the range from A to B is just [math]f_{avg}=\frac{\int^B_Af\left(x\right)dx}{B-A}[/math]. If this equation is not intuitive to you, please stare at it thoughtfully a bit more. The integral in the numerator just calculates the area under the curve. Dividing by the range will just give the average "height" of the function, which is its average value.  We can use this simple fact of calculus to find the average velocity. If we go back to the definition of displacement, we can analogously write: [br][br][center][math]\vec{v}_{avg}=\frac{\int_{t_1}^{t_2} \vec{v}\;dt}{t_2-t_1}=\frac{\vec{r}_2-\vec{r}_1}{t_2-t_1}=\frac{\Delta\vec{r}}{\Delta t}[/math].[/center]This will give the average value of velocity over the duration corresponding to the span of time between t[sub]1[/sub] and t[sub]2[/sub], which has been written [math]\Delta t[/math] here.

[i]EXAMPLE: Consider a coffee cup that starts at [/i][math]\vec{r}_i =(10\hat{i} +2\hat{j})m[/math][i] initially, and 3 seconds later is at [/i][math]\vec{r}_f =(9\hat{i}+2\hat{j})m[/math][i]. What is the average velocity during this time?[br][br]SOLUTION: First calculate [/i][math]\Delta\vec{r}[/math][i] and then divide by Δt and get [br][/i][math][br]\Delta\vec{r}=-1m\hat{i} \\[br]\vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t} \\ [br]\vec{v}_{avg}=\frac{−1}{3}\hat{i}\;m/s.[/math]

Below is an interactive graphic to explain the concept of average velocity.  Notice that the [b]magnitude of the average velocity is very different from the average speed[/b]!  The speed is the magnitude of the instantaneous velocity, but the average speed is not in general related to the magnitude of the average velocity.  You can see that following the actual path versus following the displacement generally results in the actual path being much longer, and therefore the average speed being much higher than the magnitude of the average velocity.  [br][br]In this sense, the meaning of the magnitude of the average velocity is the speed an object would have to travel at if it moves directly toward the final position (along the displacement vector) such that it arrives at the final position at the same moment as it would along the actual path.  Play with the animation to make sense of this if the words don't do it for you. 

The fact that the average velocity is the average of a vector quantity and not a scalar one means that some rather counter-intuitive things may happen. They are, however, only counter-intuitive until you understand that directions can cancel one another. Take a look at the following example that relates to the circular motion case in the animation above.

[i]EXAMPLE: Consider a 400m track and field runner who runs one lap in a time of 50s. What do you suppose their average velocity is for this single lap race?[br][br]SOLUTION: Since the beginning and ending positions are the same, the answer is [/i][math]\vec{v}_{avg}=\vec{0}\;m/s[/math][i] (a zero vector) because the displacement is a zero vector, but the average speed of the runner is [/i][math]\frac{400m}{50s}=8m/s[/math][i]. This being the case, be clear on the fact that [/i][b][i]the magnitude of the average velocity does not equal the average speed, in general[/i][/b][i]. There are exactly two instances where these two quantities are equal - when motion is in a single direction the whole time (since this constrains the motion, it can't be written as a true statement for motion in general), and when we are only considering an infinitesimal duration of time.[br] [br][/i]Given [math]\lim_{\Delta t\to 0}[/math] the magnitude of the average velocity is equal to the average speed, but we really shouldn't call this the average speed since we are only referring to an instant in time. By taking this limit we really are defining the velocity (instantaneous) once again. And we already know that the magnitude of the velocity is the speed at any instant.[br][br]​Knowing this certainly begs the question of how we define average speed correctly. Average speed, which I will denote [math]s_{avg}[/math] is just total distance covered over total time it took: [br][br][center][math]s_{avg} = \frac{total \ distance}{total \ time}.[/math][/center]Calculating the average speed for a velocity function that switches direction can be much harder than finding the average velocity.  The difference is an important one.  To find the average speed you need to know the total length of the path traveled.  This is something that is somewhat challenging to do by hand, but is trivial to do with GeoGebra. In contrast with average speed, to find the average velocity you only need to know the starting and ending points!  That is easy to do.

[i]EXAMPLE: Non-Trivial Average Speed Using GeoGebra[br][br]Consider a person jogging along a path that looks like a sine wave - like a drunk trying to walk a straight line. If the person jogs along such a path that has a 2.0 m amplitude and makes it 50 m down the sidewalk after crossing the center-line 9 times in 30 s and ending on the center line, what was their average speed?[br][br]SOLUTION: Be careful to not just reply 50/30 m/s since clearly there was additional lateral motion involved. So the distance covered was clearly greater than 50 m![br][br]We can find this path length easily within GeoGebra, but first we have to figure out the function involved that would describe the path. Clearly it's something like [math]f(x)=A\sin(kx)[/math], but we need the correct values for the constants [math]A[/math] and [math]k[/math]. The amplitude is given, so [math]A=2.0m[/math]. Since we know that the function starts and ends on the center-line and crosses it 9 times, you can draw it out and see that we get exactly 5 wavelengths to fit in 50m. This means that the function will go through a phase change of [math]5\cdot 2\pi=10\pi[/math] over 50m. So [math]10\pi=kx[/math]. This gives us [math]k=\frac{10\pi}{50m}=\frac{\pi}{5m}[/math]. Let's use GeoGebra to find the length of this path so we can finish the problem.[br][/i]