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Motion graphs and derivatives

In mechanics, the derivative of the position vs. timegraph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:

v = \frac{\Delta y}{\Delta x} = \frac{\Delta s}{\Delta t}.

Here s is the position of the object, and t is the time. Therefore, the slope of the curve gives the change in position (in metres) divided by the change in time (in seconds), which is the definition of the average velocity (in meters per second (\begin{matrix} \frac{m}{s} \end{matrix})) for that interval of time on the graph. If this interval is made to be infinitesimally small, such that {\Delta s} becomes {ds} and {\Delta t} becomes {dt}, the result is the instantaneous velocity at time t, or the derivative of the position with respect to time.

A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in y over change in x:

a = \frac{\Delta y}{\Delta x} = \frac{\Delta v}{\Delta t}.

Where v is the velocity, measured in \begin{matrix} \frac{m}{s} \end{matrix}, and t is the time measured in seconds. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives \begin{matrix} \frac{dv}{dt} \end{matrix}, the instantaneous acceleration at time t, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). The units of this slope or derivative are in meters per second per second (\begin{matrix} \frac{m}{s^2} \end{matrix}, usually termed "meters per second-squared"), and so, therefore, is the acceleration.

Since the acceleration of the object is the second derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. \begin{matrix} \frac{m}{s} \end{matrix}s = m.)

The same multiplication rule holds true for acceleration vs. time graphs. When (\begin{matrix} \frac{m}{s^2} \end{matrix}) is multiplied by time (s), velocity is obtained. (\begin{matrix} \frac{m}{s^2} \end{matrix}s = \begin{matrix} \frac{m}{s} \end{matrix}).

Variable rates of change

The expressions given above apply only when the rate of change is constant or when only the average (mean) rate of change is required. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the time-spans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are

v = \frac{ds}{dt},
a = \frac{dv}{dt}.

Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to position:

a = \frac{d^2 s}{dt^2}.

Since, for the purposes of mechanics such as this, integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:

s(t_2)-s(t_1) = \int_{t_1}^{t_2}{v}\, dt,
v(t_2)-v(t_1) = \int_{t_1}^{t_2}{a}\, dt.

Linear motion

Linear motion is motion along a straight line, and can therefore be described mathematically using only one spatial dimension. It can be uniform, that is, with constant velocity (zero acceleration), or non-uniform, that is, with a variable velocity (non-zero acceleration). The motion of a particle (a point-like object) along the line can be described by its position x, which varies with t (time). Linear motion is sometimes called rectilinear motion.

An example of linear motion is that of a ball thrown straight up and falling back straight down.

The average velocity v during a finite time span of a particle undergoing linear motion is equal to

v = \frac {\Delta d}{\Delta t}.

The instantaneous velocity of a particle in linear motion may be found by differentiating the position x with respect to the time variable t. The acceleration may be found by differentiating the velocity. By the fundamental theorem of calculus the converse is also true: to find the velocity when given the acceleration, simply integrate the acceleration with respect to time; to find displacement, simply integrate the velocity with respect to time.

This can be demonstrated graphically. The gradient of a line on the displacement time graph represents the velocity. The gradient of the velocity time graph gives the acceleration while the area under the velocity time graph gives the displacement. The area under an acceleration time graph gives the velocity.

Linear motion is the most basic of all motions. According to Newton's first law of motion, objects not subjected to forces will continue to move uniformly in a straight line indefinitely. Under every-day circumstances, external forces such as gravity and friction will cause objects to deviate from linear motion and can cause them to come to a rest.

For linear motion embedded in a higher-dimensional space, the velocity and acceleration should be described as vectors, made up of two parts: magnitude and direction. The direction part of these vectors is the same and is constant for linear motion, and only for linear motion

From Yahoo Answers

Question:I cant derive 3rd equation of motion i.e. V raise to power 2= U Raise to power 2 +2AS

Answers:Which graph du you use? velocity verses time or velocity versus distance?

Question:We only used 3 equations so far. v = u + at v = u + 2as s = ut + at Anyway, out teacher told us that the fourth equation doesn't work because there is a variable that we need to take into consideration that we don't know about. Bear in mind that we are doing all this work a entire year ahead of the syllabus. The equation is s = (u+v)/2 * t What is that variable?

Answers:The first three equations are for constant acceleration and for displacement at t=0 equal to zero, s(0)=0. What you will learn as you progress is that Sir Issac Newton invented calculus as the mathematical structure of classical physics. THese equations are related to the integrals and derivatives of each other. The variable missing in the fourth equation is the displacement at t=0, so one would wrote s=s0+(u+v)/2 * t also, the general equation is s(t)=s0=ut+1/2at^2 ___________________-

Question:What is the basic difference in quality and type of results when solving the same system of equations by the algebraic method and the graphical method? Illustrate your answer with examples.

Answers:well let's take y=x^2 - 2 for example algebraic way will give us exact roots where graphical we would have to trace the graph to find these roots sometimes x^2 - 2 can be factored as difference of two squares as (x+sqrt(2))(x-sqrt(2) as we all know sqrt(2) is irrational, but to say that x = + - sqrt(2) is a lot more accurate than guessing. that's my two cents. The graphical method is nice because you don't have to do a whole lot with the equation to see that the curve looks like. Algebtaicly you need to take the first derivative to see what the slope does (increase/decrease) and the second der. to find inflection points. You may also need to factor the equation to find asymptots... but plugging it into the calculator and graphing it is a whole lot easier.

Question:Plz could any one send me a link for most precise derivation of "Tangential acceleration in circular motion"? I need a link to Derivation of tangential acceleration.....its formulas and equations....also the derivation of velocity and acceleration vectors......

Answers:tangential acc is angular tip speed acceleration. angular acceleration (alpha) rad/sec x radius = tip acceleration m/s^2

From Youtube

How to Derive Equations of Motion - With and Without Calculus :Here's how to derive the 4 major kinematic equations (aka: equations of motion) that are used to describe the motion of particles in physics. I show how to derive the both with and without calculus. These equations are valid only if accleration is constant. Acceleration can be equal to zero, however it can't change. If you have a changing acceleration, then calculus must be used to get a different set of equations. www.alexpleasehelp.com

Mod-5 Lec-1 Generalized and Principle Coordinates, Derivation of Equation of Motion :Lecture Series on Mechanical Vibrations by Prof.SKDwivedy, Department of Mechanical Engineering, IIT Guwahati. For more details on NPTEL visit nptel.iitm.ac.in