If we take the derivative of velocity with respect to time we get the acceleration of an object. The acceleration must therefore tell us the rate at which the velocity is changing. Keep in mind that the velocity has both a magnitude and a direction. If either one of those changes, there will be a non-zero acceleration. Since velocity is the time derivative of position, we may write acceleration in terms of either velocity or position:[br][br][center][math]\vec{a}=\frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}[/math].[/center]The units of acceleration are m/s[sup]2[/sup] since it is the rate of change of the rate of change of position.[br][br][i]A NOTE ON UNITS: In old textbooks, the unit of acceleration was often written m/s/s. This would be read as meter per second per second. While this doesn't look pretty to modern eyes where equations are nicely typeset, it is perhaps more useful in understanding what acceleration implies than our modern version. When you see [math]a=10\frac{m}{s^2}[/math], it really means that v is changing by 10 m/s every second, or 10 m/s/s as it was written in old textbooks.[/i]

In the same way that we can integrate velocity with respect to time to arrive at displacement, we can also integrate acceleration with respect to time to get the change in velocity over the time interval of the integration. This follows from the same mathematics as before:[br][br][center][math][br]\vec{a}=\frac{d\vec{v}}{dt} \cr[br]\vec{a}\;dt=d\vec{v} \cr[br]\int_{t_1}^{t_2}\vec{a}\;dt=\int d\vec{v} \cr[br]\int_{t_1}^{t_2}\vec{a}\;dt=\vec{v}_2-\vec{v}_1=\Delta\vec{v}.[br][/math][/center][br]This means that we can find the later velocity from the earlier one by just adding the change, or [math]\vec{v}_2 = \vec{v}_1+\Delta\vec{v}[/math]. Furthermore, now we can also continue on and find the displacement and the final position. The displacement is [math]\Delta\vec{r}=\int \vec{v}\;dt = \int (\vec{v}_1+\Delta\vec{v})\;dt[/math]. Thus the final position is [math]\vec{r}_2=\vec{r}_1+\Delta{\vec{r}}[/math]. In both of these equations, the terms with the [math]\Delta s[/math] are to be arrived at by evaluating the integrals.

One common example of acceleration is the category of problem that deals with objects falling under the influence of gravity near the earth's surface while neglecting other forces.  It turns out that all objects subject to gravity alone accelerate at the same rate, for reasons that will become clear in the chapter on Newton's laws.  When such objects are dropped, they accelerate at approximately 10 m/s[sup]2[/sup], independent of their mass.  The actual value is 9.81m/s[sup]2[/sup].  This term is denoted [math]\vec{g}.[/math]  Notice that this is a vector.  It has a direction which is toward earth's center. I prefer you think of this not as an acceleration, but as a gravitational field strength, but the reality is that the field strength carries the same units as acceleration and so it is often called acceleration.[br][br]The reason it's better to think of it as field strength is that gravity pulls on objects near earth's surface whether they are falling or not. Therefore [math]\vec{g}[/math] is present even apart from any consequent acceleration. I believe calling the gravitational field strength an acceleration is the root of many misconceptions and confusions among students.  [br][br]If we expect objects to fall and accelerate at a rate [math]\vec{a}=\vec{g}[/math], we must neglect all other forces involved.  The most common of these neglected forces is air drag.  Soon we will learn how to handle such problems without neglecting forces such as air drag, and it will be done by means of numerical methods.  While you might think it's silly or careless to use the approximation (10 m/s[sup]2[/sup]) for the gravitational field strength, consider the fact that the bigger approximation is the neglect of air drag most of the time.  So when we do problems with the gravitational field that neglect drag we will use 10 m/s[sup]2[/sup].  On the other hand, if we take the time to carefully calculate the effects of air drag, then we will use the more precise value for the gravitational field.  Furthermore, the approximation is only off by 2%, which is better than we'll manage to measure most quantities in lab.

[i]QUESTION: A rock is dropped from a 50m tall cliff.  How long will it take to reach the valley below?  (Assume air drag is negligible.)[br][br]SOLUTION: Out of convenience, we can assume that we start from the origin and from rest.  The acceleration will be uniformly downward at 10 m/s[sup]2[/sup] for the entire fall since we are neglecting air drag.[br][br]We can arrange our axes however we want.  Let's just use the x-coordinate and point it downward so that all variables are positive valued as the rock falls.  This way we only need to keep track of the x-components of all our vectors.[br][br]We may start by finding the change in velocity, or [/i][math]\Delta v_x=\int a_xdt[/math][i]. This gives [math]v_x=v_{0x}+a_x t=a_x t,[/math] since the initial velocity is zero. Next we need to use this to find displacement. [math]\Delta r_x=\int a_x t\;dt=a_x\frac{t^2}{2}[/math][/i]. [i]Inserting numbers that were given leads to t=3.2s by using 50m for the displacement.[br]  [br]QUESTION: How fast will this same rock be traveling the instant before it strikes the ground?[br]SOLUTION: Using the first part of the work above will lead to the answer of [math]v_x=32m/s[/math][br]by simply plugging in the time we just found.[br][/i]

Separating the magnitude from the direction of a vector can be very useful. Presently we will see one very useful situation. Suppose an object is moving along at some known speed v and we want to deflect its path and change its direction of travel while leaving its speed constant. The object in question could be anything from a car traveling around a bend to an asteroid floating in space - that part is irrelevant. Here is how the math works: [br][br][center][math]\vec{a}=\frac{d\vec{v}}{dt} \\[br]\vec{a}=\frac{d}{dt}(v\hat{v}) \\[br]\text{Use the product rule of differentiation:} \\[br]\vec{a}=\frac{dv}{dt}\hat{v} + v\frac{d\hat{v}}{dt}. [br][/math][/center][br][br]The first term describes a change in speed, which is what most people think of when they hear the word acceleration. We can see from the math that when speed changes, the acceleration points along the velocity vector, or in the direction of [math]\hat{v}.[/math] Take note that if an object is slowing down, so that [math]\tfrac{dv}{dt}[/math] is negative, that the acceleration will point in opposition to the velocity's direction. That means a car driving northward while slowing down is accelerating southward and an elevator traveling upward while slowing down is accelerating downward.[br][br]The second term above in the acceleration relates to a change in direction while traveling at a fixed speed v. The term [math]\tfrac{d\hat{v}}{dt}[/math] describes the rate of change of the direction. An obvious choice for units would be the amount of angle by which the direction changes divided by time. The only caveat is that a unit vector is unitless, so the angle unit must not be a unit. Here radians come to our rescue since they are really a ratio and not a unit (contrasted with degrees). So that term is described by the number of radians per second by which the path is being changed or redirected. [br][br]What does this mean? Suppose you want to know how much acceleration you experience as you make a U-turn at 10 m/s on a bicycle during a five second duration of time. Since a U-turn is 180 degrees, or pi radians (we need radians), this gives [math]\vec{a}=10m/s\frac{\pi rad}{5s} = 2\pi \tfrac{m}{s^2}.[/math] Notice that the radians just vanished. They do that since they aren't real units. Notice a problem here? We lost the direction. Let's discuss that next. [br][br]A picture of two position vectors - the earlier position [math]\vec{r}_1[/math] and later position [math]\vec{r}_2[/math] - of an object as it travels along a curve. The unit velocity vectors are also shown. To the right, the two unit velocity vectors are shown tail to tail so that we can see what changed. The change (which is what we're after) is shown. Notice that it points toward the center of the curve along which the object is traveling. Since this is inward in the radial sense, we call this the [math]-\hat{r}[/math] direction. Negative means inward and 'r' stands for radial. So the acceleration above can be written then as [math]\vec{a}=-2\pi \tfrac{m}{s^2}\hat{r}.[/math] The negative goes with the unit vector to describe direction of the acceleration as discussed.