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Equations of one-dimensional motion

Classical mechanics

Equations of one-dimensional motion

We discuss velocity, acceleration, displacement etc. and the three important equations of motion — including both the arithmetic and calculus derivations.

Our discussion so far has been with regard to dynamics using Newton’s laws: the ideas of forces and their effects on bodies. Today we will talk about the mathematics driving a particular kind of motion — in a straight line — and concerning only the displacement (S), velocities (uv), acceleration (a) and time (t). These are called the famous SUVAT equations, or the equations of one-dimensional motion.

These form part of the concept of studying the behaviour of bodies by reducing them to mathematical models: kinematics.

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Definitions

Before we dive into the actual equations themselves, let us define the SUVAT terms we will be using.

0. Position

We shall call position our zeroth definition because we do not need it fundamentally in our coming equations.

Position of an object is whatever you determine it to be. An object is somewhere in space, and you can locate it by placing a co-ordinate system around it, with the origin at any point, with the axes at any tilt and so on. It is alright if you do not understand this idea fully right now, but know that an object’s position answers the question “where in space is it?”.

For our discussion today, we will confine positions to one dimension and refer to it using the term x.

1. Displacement

This is the shortest (in our case straight line) distance between two points that an object can take.

For instance, in fig. 1, one could go from A to B along these two routes. Here, both routes 1 and 2 are the distances between A and B, but the shortest possible route (route 1) is given the special name of displacement.

displacement or distance

Fig. 1

A more accurate definition of displacement would be to call it the only route between two points with constant direction. Hence, displacement is a vector, while distance is not.

Note that we could, for convenience, call the object’s position at A as x_A and at B as x_B and so on.

2. Velocity

Velocity is the vector form of speed. If speed is the rate at which an object moves, or the change of an object’s position with respect to time, velocity is speed in a particular direction.

Using the help of our definition of displacement above, we might say, if an object maintains a speed of 60kmph on route 2, it has a speed of 60kmph. If it maintains some 80kmph on route 1, then it still has a speed of 80kmph, but this speed with a constant direction as in route 1 is given the special term velocity.

Velocity, therefore, is the rate of change of an object’s position in a particular direction. This makes velocity a vector.

It is worth noting that this is a rather unrealistic depiction. Objects, say your bike, hardly maintain the same speed throughout a journey, instead changing at every interval of time, however small that interval may be. This should mean that realistically velocities never exist, but we consider what is termed instantaneous velocity, the velocity of an object at any split-second point in time.

We represent the velocity of an object at clock start (t=0, see below with the definition of time) using u, the initial velocity. Its velocity at the final moment t_{f} is called the final velocity, represented by v.

For those interested in a more mathematical definition, \mathbf{v}=\frac{dx}{dt} at any time t when the object is at position x. This is written in short as \mathbf{v}=\dot{x}(t).

In the simplest of terms,

    \begin{align*} velocity=\frac{displacement}{time} \end{align*}

3. Acceleration

If velocity is the rate of change of position of an object in a given direction, then the rate of change of its velocity over several time intervals is called its acceleration.

Note that having said rate of change of velocity, we have already introduced the idea of a specific direction. Hence acceleration also has a specific direction and is a vector.

Acceleration is represented by a and mathematically is \mathbf{a}=\frac{d\mathbf{v}}{dt} which is also \mathbf{a}=\frac{d\frac{dx}{dt}}{dt} or simply \mathbf{a}=\ddot{x}(t)=\dot{v}(t).

Once again, in simple terms,

    \begin{align*} acceleration=\frac{velocity}{time} \end{align*}

What is an equation of motion?

Generally, any equation that establishes a relation between the properties of displacement, velocities, acceleration and time comparisons of an object are called its equations of motion. That is not to say an equation of motion includes all of these, but it often relates at least any two of them.

A fancy way of saying this is that for any equation f to be an equation of motion, it is to some degree of the form f \left[ t, \mathbf{x}(t), \dot{\mathbf{x}}(t), \ddot{\mathbf{x}}(t) \right] which simply means that f must contain any of time (t), position (\mathbf{x}(t)), velocity (\dot{\mathbf{x}}(t)) and/or acceleration (\ddot{\mathbf{x}}(t)). The dots above x suggest derivatives with respect to time: one dot is \frac{d}{dt}, two are \frac{d\frac{d}{dt}}{dt} and so on.

The equations of motion

The next part of our discussion will involve small derivations of equations of motion. Derivations are meant to convince us of the truth in a useful final result. You can follow them if you are interested or simply look at eqn. (1) to (3) by skipping any derivations (i.e. look at the final equations in the boxes; these are the equations of motion).

Remember that equations of motion are also known as kinematics equations. Once again, we will concern ourselves only with one dimension. (Note: a calculus-based set of derivations follow this arithmetic derivation after this section.)

velocity time graph

Fig. 2

Let us say we have an object moving with increasing velocity. That means the object has some acceleration. We might represent it on a simple graph like the on in fig. 2. We know that the acceleration (see the simple definition above) can be represented over some change in velocity \Delta \mathbf{v} and time t as,

    \begin{align*} \mathbf{a} & = \frac{ \Delta \mathbf{v} }{ t } \ \implies \mathbf{a} & = \frac{ \mathbf{v} - \mathbf{u} }{ t } \end{align*}

(1)   \begin{align*}  \therefore \boxed{ \mathbf{ v } = \mathbf{u} + \mathbf{a} t } \end{align*}

Let us consult fig. 2 once again at this point. The area under the line of the graph i.e. the area OABCD gives the displacement of the object. This is a general observation: on velocity-time graphs, the slope gives the acceleration, the area gives the displacement.

To solve the area mathematically, observe how OABCD forms the closed figures ABC (a triangle) and ACDO (a rectangle). The areas then are respectively A_{triangle} = \frac{1}{2} \times base \times height which comes to \frac{1}{2} \times t \times (\mathbf{v}-\mathbf{u}) and A_{rectangle}=length \times breadth which is \mathbf{u} \times t.

We know that t \times (\mathbf{v}-\mathbf{u}) = t \times ( ( \mathbf{u} + \mathbf{a} t ) - \mathbf{u}) from eqn. (1) which reduces to A_{triangle}=\frac{1}{2} \times \mathbf{a} t^2. Summing up the two areas gives us the total area, i.e. the displacement S:

(2)   \begin{align*}  \boxed{ \mathbf{S}  = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 } \end{align*}

Our fundamental derivations end at these equations for now. We will use eqn. (1) and (2) to derive the third equation of motion. Let us re-write eqn. (1) in terms of t as,

(1b)   \begin{align*}  t = \frac{ \mathbf{v} - \mathbf{u} }{ a } \end{align*}

And now we substitute eqn. (1b) in (2),

    \begin{align*} \mathbf{S}  = \mathbf{u} \left( \frac{ \mathbf{v} - \mathbf{u} }{ a } \right) + \frac{1}{2} \mathbf{a} \left( \frac{ \mathbf{v} - \mathbf{u} }{ a } \right)^2 \end{align*}

Cancelling \mahtbf{a^2} and a in the second term on the right hand side, and moving 2 and the common a denominator onto the left hand side,

    \begin{align*} 2 \mathbf{a} \mathbf{S} & = 2 \mathbf{v} \mathbf{u} - 2 \mathbf{u^2} + \left( \mathbf{v} - \mathbf{u} \right)^2 \ & = 2 \mathbf{v} \mathbf{u} - 2 \mathbf{u^2} + \mathbf{v^2} + \mathbf{u^2} - 2 \mathbf{v} \mathbf{u} \ & = \mathbf{v^2} - \mathbf{u^2} \end{align*}

Re-arranging terms,

(3)   \begin{align*}  \therefore \boxed{ \mathbf{v^2} = \mathbf{u^2} - 2 \mathbf{a} \mathbf{S} } \end{align*}

Equations (1) to (3) are our three equations of motion used only under circumstances where the acceleration is constant. This is quite obvious from the graph we used right in the beginning. The velocity-time graph is an oblique but straight line, signifying constant acceleration.

That is all for our basic equations of one-dimensional motion. If you are interested in a calculus-based derivation, click below to reveal the next (optional) section.

OPTIONAL SECTION

The calculus derivations

We like to think of this next set of derivations as a fine example of why calculus is used predominantly in physics. It might seem hard, but calculus radically simplifies our derivations.

We will skip specific details for now as this next section is targeted at readers already familiar with calculus. In each case we integrate instantaneous values over the entire 0 to t time period (where t is present) or u to v velocities (where velocity is present).

Also, since these apply when acceleration is constant, we treat a as such when integrating or differentiating, often keeping it as it was.

Equation 1

    \begin{align*} \mathbf{a} & = \frac{d \mathbf{v}}{dt} \\\ \mathbf{a} dt & = d \mathbf{v} \\\ \int_0^t \mathbf{a} dt & = \int_u^v d \mathbf{v} \\\ \mathbf{a} t & = \mathbf{v} - \mathbf{u} \\\ \end{align*}

thus,

(1)   \begin{align*}  \boxed{ \mathbf{v} = \mathbf{u} + \mathbf{a} t } \end{align*}

Equation 2

    \begin{align*} \mathbf{v} & = \frac{dx}{dt} \\\ \mathbf{v} dt & = dx \\\ \left( \mathbf{u} + \mathbf{a} t \right) dt & = dx \\\ \int_0^t \left( \mathbf{u} + \mathbf{a} t \right) dt & = \int_0^S dx \\\ \mathbf{u} t + \frac{\mathbf{a} t^2}{2} & = \mathbf{S} \\\ \end{align*}

Re-arranging terms,

(2)   \begin{align*}  \boxed{ \mathbf{S} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 } \end{align*}

Equation 3

This one requires a little trick right at the start (see chain rule after the derivation):

    \begin{align*} \frac{d \mathbf{v} }{dx} & = \frac{d \mathbf{v} }{dx} \frac{dt}{dt} \\\ & = \frac{d \mathbf{v} }{dt} \frac{dt}{dx} \\\ & = \mathbf{a} \frac{1}{\mathbf{v}} \\\ \mathbf{v} \; d \mathbf{v} & = \mathbf{a} dx \\\ \int_u^v \mathbf{v} \; d \mathbf{v} & = \int_0^S \mathbf{a} dx \\\ \left.\frac{v^2}{2} \right\vert_u^v & = \mathbf{a} \left( \mathbf{S} \right) \\\ \frac{1}{2} \left( \mathbf{v^2} - \mathbf{u^2} \right) & = \mathbf{a} ( \mathbf{S} ) \end{align*}

which reduces to,

(3)   \begin{align*}  \boxed{ \mathbf{v^2} = \mathbf{u^2} + 2 \mathbf{a} \mathbf{S} } \end{align*}

Chain rule

Thanks to Subramanya for pointing out below (see comments) that it is worth noting that this trick is really a basic method of calculus known as the chain rule. The rule simply provides a solution when y = f(w) and w = g(x) when there is no direct way of writing the derivative of y with respect to x (i.e. \frac{dy}{dx}) because y is given with respect to w and not x. However, we know x in terms of something else (g(x)) which we can use to compute \frac{dy}{dx} using the chain rule, which says, \frac{dy}{dx} = \frac{dy}{dw} \frac{dw}{dx}.

How does this work? The more general statement of the chain rule for every F(x) = (f \circ g) (x) goes,

    \begin{align*} F'(x) = f'(g(x))g'(x) \end{align*}

The meaning of (f \circ g) is simply f(g(x)). Comparing it to our example above, note how, if y = f(w) and w = g(x), then y=f(g(x)). In derivation 3 above, this is precisely what we apply. We use \frac{dt}{dt} because it effectively reduces to 1 (multiplying by 1 makes no known difference) in order to bring it to the form \frac{dy}{dw} \frac{dw}{dx}.

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V.H. Belvadi is an Assistant Professor of Physics. He teaches postgraduate courses in advanced classical mechanics, astrophysics and general relativity. When he is free he makes photographs and short films, writes on his personal website, makes music, reads voraciously, or plays his violin. He currently serves as the Editor-in-Chief of Physics Capsule.

3 Comments

3 Comments

  1. Subramanya Hegde

    August 2, 2014 at 9:35 am

    It would have been better if you had explicitly mentioned the steps in which you were using constancy of acceleration. And it is better to mention that you are using the chain rule of differentiation rather than speaking of a ‘trick’ – multiplying and dividing by dt.

    • V.H. Belvadi

      August 2, 2014 at 10:03 am

      Thank you, Subramanya.
      I did not want to make the calculus overbearing with rules, but what you make is a valid point about it not being a “trick” entirely.
      I have updated the article with a brief word on the chain rule.

  2. Pingback: The mass-spring system - part 3

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