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Equations of Motion

Cut to the Chase


These equations of motion are valid only when acceleration is constant and motion is constrained to a straight line. In the real world this is an unrealistic notion. No object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time. However it would be wrong to dismiss this section outright or as useless.

In many instances, it is useful to assume that an object did or will travel along a path that is essentially straight and with an acceleration that is nearly constant. That is, any deviation from the ideal motion can be essentially ignored. Motion along a curved path may also be effectively one-dimensional if there is only one degree of freedom for the objects involved. For example, a road might twist and turn and explore all sorts of directions, but the cars driving on it have only one degree of freedom: the freedom to drive in one direction or the opposite direction. (You can't drive diagonally on a road and hope to stay on it for very long.) In this regard, it is not unlike motion restricted to a straight line. Approximating real situations with models based on ideal situations is not considered cheating. This is the way things get done in physics. It is such a useful technique that we will use it over and over again.

Our goal in this section, is to derive the equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity, displacement, and time. There are three ways to pair them up: velocity-time, displacement-time, and velocity-displacement. In this order, they are also often called the first, second, and third equations of motion, but there is no compelling reason to learn these names. Since we are dealing with motion in a straight line, the symbol x will be used for displacement. The direction of motion will be indicated by the sign (positive quantities point in +x direction, while negative quantities point in the -x direction). Determining which direction is positive and which negative is entirely arbitrary. The laws of physics are isotropic; that is, they are independent of the orientation of the coordinate system. As long as you are consistent, it doesn't matter. Some problems are easier to understand and solve, however, when one direction is chosen positive over another.


The relation between velocity and time is a simple one during constantly accelerated, straight-line motion. Constant acceleration implies a uniform rate of change in the velocity. The longer the acceleration, the greater the change in velocity. If after a time velocity increases by a certain amount, after twice that time it should increase by twice that amount. Change in velocity is directly proportional to time when acceleration is constant. If an object already started with a certain velocity, then its new velocity would be the old velocity plus this change. You ought to be able to see the equation in your mind's eye already.

This is the easiest of the three equations to derive formally. Start from the definition of acceleration, expand the Δv term, and solve for v as a function of t.


Since acceleration is also the first derivative of velocity with respect to time, this equation can also be derived using calculus. Just reverse the action of the definition. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. Since acceleration is assumed constant, this is quite easy.

dv=a dt
\int^{v}_{v_0}dv=\int^{\Delta{t}}_{0}a dt

The symbol v0 (v nought) is called the initial velocity. It is also given the symbol u in many texts. The initial velocity is the velocity that a moving object has when it first becomes important in a problem. For example, say a meteor was spotted deep in space and the problem was to determine its trajectory, then the initial velocity would be the velocity at the time it was observed. But if the problem were to determine its velocity on impact, then it's initial velocity would more likely be the velocity it had when it entered the earth's atmosphere. In this case, the answer to, "What's the initial velocity?" is "It depends". This turns out to be the answer to a lot of questions.

The symbol v is then the velocity some time Δt after the initial velocity is often called the final velocity. What is taken to be the final velocity is depends on the problem you are solving. There is no hard and fast rule.

The last part of this equation aΔt is the change in the velocity from the initial value. Recall that a is the rate of change of velocity and that Δt is the time interval since the object had its initial velocity v0. Rate multiplied by time equals change. Thus if an object were accelerating at 10 ms-2, after 5 s it would be moving 50 ms-1 faster than it was initially. If it started with a velocity of 15 ms-1, its velocity after 5 s of acceleration would be 65 ms-1.


The displacement of a moving object is directly proportional to both velocity and time. Acceleration compounds this simple situation. Now the velocity is also directly proportional to time. Try saying this in words and it sounds ridiculous. "Displacement is directly proportional to time and directly proportional to velocity, which is directly proportional to time." Time is a factor twice, making displacement proportional to the square of time. A car accelerating for two seconds would cover four times the distance of a car accelerating for only one second (22 = 4). A car accelerating for three seconds would cover nine times the distance (32 = 9).

Would that it should be so simple. This example only works when initial velocity is zero. Change in displacement is proportional to the square of time when acceleration is constant and initial velocity is zero. A true general statement would have to take into account any initial velocity and how the velocity was changing. This results in a terribly messy proportionality statement. Change in displacement is directly proportional to time and proportional to the square of time when acceleration is constant. A function that is both linear and square is said to be quadratic, which allows us to compact the previous statement considerably. Change in displacement is a quadratic function of time when acceleration is constant

Proportionality statements are useful, but not as concise as equations. We still don't know what the constants of proportionality are for this problem. The only way to answer that is through algebra.

Start with the definition of velocity, expand Δx, and solve it for displacement.



To continue, we need to resort to a little trick first published in the Fourteenth Century at Merton College, Oxford (and sometimes called the Merton Rule). When acceleration is constant, the velocity will change uniformly from its initial value to its final value and the average will lie halfway between the extremes. Thus, the average velocity is just the arithmetic mean of the initial and final velocities. Average velocity is the average of the final and initial velocities when acceleration is constant.


Substitute the first equation of motion [1] into this equation (4) and simplify with the intent of eliminating v.



Finally, substitute [b] into [a] and solve for x as a function of t.



Since velocity is also the first derivative of displacement with respect to time, this equation can also be derived using calculus. In fact, it's much easier than using algebra. Just reverse the action of the definition. Instead of differentiating displacement to find velocity, integrate velocity [1] to find displacement.



The symbol x0 (ex nought) is the initial displacement. Many times, this value is zero and if it isn't, we can make it so. If you ask me, "When should we do this?" I would say, "It depends on the problem," and leave it to you to decide. There is no rule that you can memorize is this case. You have to understand what the equation says and then learn how to apply it to a particular situation. Similarly x is often called the final displacement, but this does not make it the "last displacement", rather it is the displacement at the end of the time interval during which the acceleration was constant.

Something else to notice is the similarity between equations [2] and [a]. When acceleration is zero, our second equation of motion reverts to a rearranged constant velocity equation. As predicted, displacement is in part directly proportional to time and in part directly proportional to time squared.





Although the velocity symbols in the two equations may look different, they do indeed represent the same quantity. If there is no acceleration, then the velocity is constant, which means that the initial velocity is the same as the final velocity is the same as the average velocity. The acceleration term at the end is an adjustment to the constant velocity equation to account for the the fact that the velocity is changing. A positive acceleration would increase the displacement and a negative acceleration would decrease it. This is exactly what one would expect. If an object's velocity was increasing, it would move farther than if it had stayed at a constant velocity. Likewise, if an object's velocity was decreasing, it would have a smaller displacement than if its velocity were constant. It's good to see that the equations behave in a realistic manner. Otherwise all this math would be a waste of time.


We have just seen that velocity is directly proportional to time and displacement is proportional to time squared. With a little bit of thinking, a new proportionality statement should be apparent. Change in displacement is proportional to the change in the square of velocity when acceleration is constant. This statement is particularly important for driving safety. When you double the speed of a car, it takes four times the distance to stop it. Triple the speed and you'll need nine times the distance. Like the previous relationships, it also depends on initial velocity. Unfortunately determining the effect using reasoning alone is a real chore. Do the algebra instead.

The last two equations each described one kinematic variable as a function of time. It would be nice if we also had an equation that was independent of time. That is, we want to answer the question, "What is the relationship between velocity and displacement?" The method of doing this should be readily apparent. We've got to combine our first two equations of motion together in a manner that will eliminate time as a variable. The easiest way to do that is to solve one equation for time and then substitute it into the other. The second equation of motion is a quadratic and solving it for time would introduce a lot of nastiness into the algebra. It should be apparent that solving the first equation of motion [1] for time and substituting it into the second [2] will be the easier process.


Substituting Δt=(v-v0)/a



As expected, displacement is proportional to velocity squared. (Initial velocity squared is just a constant that must be dealt with.) Unlike the first and second equations of motion, there is no advantage in using calculus to derive the third equation of motion. Algebra is the way to go.