The Physics
Hypertextbook
Opus in profectus

# Equations of Motion

## Discussion

### constant acceleration

In order to be accurate, the title of this section should be "One Dimensional Equations of Motion for Constant Acceleration". Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. These equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.

Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time — not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute metaphysical certainty.

So what good is this section then? Well, 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. 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 new 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. Direction 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.

### velocity-time

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.

 a = Δv = v − v0 Δt Δt v = v0 + aΔt [1]

The symbol v0 [v nought] is called the initial velocity. It is often thought of as the "first velocity" but this is a rather naïve way to describe it. Take the case of a meteor hurtling towards the Earth. What is its initial velocity? If you want v0 to be the first velocity, then the problem is over before it started. Who could possibly say what a meteor's velocity was at its birth? There is no way to answer this question. A better definition would be to say that an initial velocity is the velocity that a moving object has when it first becomes important in a problem. Say the 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. It is often called the final velocity but this does not make it an object's "last velocity". Take the case of the meteor. What velocity is represented by the symbol v? If you've been paying attention, then you should have anticipated the answer. It depends. It could be the velocity of the meteor as it passes by the moon, as it enters the Earth's atmosphere, or as it strikes the Earth's surface. It could also be the meteorite's velocity as it sits in the bottom of a crater (zero). Is this then the final velocity? Who knows. Someone could extract the meteorite from its hole in the ground and drive away with it. Is this relevant? Well, maybe. It depends. There's no rule for this kind of thing. You'll have to understand the problem and then make a decision for yourself.

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 m/s2, after 5 s it would be moving 50 m/s faster than it was initially. If it started with a velocity of 15 m/s, its velocity after 5 s of acceleration would be…

15 m/s + 50 m/s = 65 m/s.

### displacement-time

The displacement of a moving object is directly proportional to both velocity and time. Move faster. Go farther. Move longer (as in longer time). Go farther. 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.

 v̅ = Δx Δt Δx = v̅Δt x − x0 = v̅Δt x = x0 + v̅Δt [a]

To continue, we need to resort to a little trick first published in 1335 by the English philosopher, mathematician, logician, and "calculator" William Heytesbury (1313–1372) known as the mean speed rule. Heytesbury was a fellow at Merton College, Oxford so this trick is also known as the Merton rule (the name I prefer since the rule applies to quantities other than speed — as long as they change at a uniform rate). 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. In other words, average velocity is just the average of the initial and final velocities.

 v̅ = ½(v + v0) [4]

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

 v̅ = ½(v + v0) v̅ = ½[(v0 + aΔt) + v0] v̅ = ½(2v0 + aΔt) v̅ = v0 + ½aΔt [b]

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

 x = x0 + v̅Δt x = x0 + (v0 + ½aΔt)Δt x = x0 + v0Δt + ½aΔt2 [2]

The symbol x0 [x 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 in 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.

 x = x0 + v0Δt + ½aΔt2 [2] x = x0 + v̅Δt [a]

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.

### velocity-displacement

The first two equations of motion each describe one kinematic variable as a function of time. In essence…

1. Velocity is directly proportional to time when acceleration is constant (v ∝ t).
2. Displacement is proportional to time squared when acceleration is constant (s ∝ t2).

Combining these two statements gives rise to a third — one that is independent of time. With a little bit of thinking, it should be apparent that…

1. Displacement is proportional to velocity squared when acceleration is constant (s ∝ v2).

This statement is particularly relevant to driving safety. When you double the speed of a car, it takes four times more distance to stop it. Triple the speed and you'll need nine times more distance. This is a good rule of thumb to remember.

The conceptual introduction is done. Time to derive the formal equation. The method of doing this should be apparent. Combine the first two equations together in a manner that will eliminate time as a variable. The easiest way to do this is to solve the first equation [1] for time…

 v = v0 + aΔt [1] Δt = v − v0 a

and substitute it into the second equation [2]…

 x = x0 + v0Δt + ½aΔt2 [2] x − x0 = v0Δt + ½aΔt2

like so…

 x − x0 = v0 ⎛⎝ v − v0 ⎞⎠ + 1 a ⎛⎝ v − v0 ⎞2⎠ a 2 a
 x − x0 = vv0 − v02 + v2 − 2vv0 + v02 a 2a
 2a(x − x0) = 2(vv0 − v02) + (v2 − 2vv0 + v02)
 2a(x − x0) = v2 − v02
 v2 = v02 + 2a(x − x0) [3]

That wasn't very pleasant, but it was the easy way. The reverse substitution of [2] into [1] is an algebraic nightmare.

### calculus derivations

Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity–time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].

 a = dv dt dv = a dt v Δt ⌠⌡ dv = ⌠⌡ a dt v0 0 v − v0 = aΔt v = v0 + aΔt [1]

Again by definition, velocity is the first derivative of displacement with respect to time. Reverse the operation in the definition. Instead of differentiating displacement to find velocity, integrate velocity to find displacement. This gives us the displacement–time equation for constant acceleration, also known as the second equation of motion [2].

 v = dx dt dx = v dt = (v0 + at) dt x Δt ⌠⌡ dx = ⌠⌡ (v0 + at) dt x0 0 x − x0 = v0Δt + ½aΔt2 x = x0 + v0Δt + ½aΔt2 [2]

Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to displacement) using calculus. We can't just reverse engineer the definitions. We need to play a rather sophisticated trick.

The first equation of motion relates velocity to time. We essentially derived it from this derivative…

 dv = a dt

The second equation of motion relates displacement to time. It came from this derivative…

 d2x = a dt2

The third equation of motion relates velocity to displacement. By logical extension, it should come from a derivative that looks like this…

 dv = ? dx

But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1, dt/dt, and then do a bit of special algebra — algebra with infinitesimals. Look what happens when we do this. We get a derivative equal to acceleration and another equal to the inverse of velocity.

 dv = dv · dt = dv · dt = a 1 dx dx dt dt dx v

Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…

 dv = a 1 dx v v dv = a dx v x ⌠⌡ v dv = ⌠⌡ a dx v0 x0 ½(v2 − v02) = a(x − x0) v2 = v02 + 2a(x − x0)

Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However…. It really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving.

 v = v0 + aΔt [1] + x = x0 + v0Δt + ½aΔt2 [2] = v2 = v02 + 2a(x − x0) [3]