[Following is a text which assumes that the reader is familiar with basics of both Calculus and Vectors.Are Calculus and Vectors important?Believe me if you have good hold on both Calculus and Vectors, all you need is a good book to understand Physics.You can do most of the things yourself]

Kinematics involves problems concerning motion of objects.But, in kinematics we don't take into account the reason for variation in the motion,i.e,we don't use the concepts of Force and Mass to determine the motion.Then how is it determined?

You can know about the motion of a particle by only knowing its position **(r)** with respect to time **(t)**.If **'r'** denotes the position of a particle and **'t'** denotes time, then, surely **'r'** is a function of **'t'**,i.e,**'r'** depends on **'t'**.Notice that **'r'** is a vector quantity whereas**'t'** is a scalar quantity.

Now,imagine a physical system consisting of a single particle.The position of the particle with respect to a reference frame, at an arbitrary moment in time **'t'**, is determined by the position vector**(r)** [or radius vector] of the particle. If we introduce the unit vectors **i,j,k** along the **x,y,z** axes, respectively, the radius vector **r** can be represented in the following form

\[r(t)=x(t)\hat { i } +y(t)\hat { j } +z(t)\hat { k }\quad ...(1.1)\]

with **x(t)**, **y(t)** and **z(t)** as the projection of **r(t)** along **x,y and z** axis, respectively.Simultaneously specifying three functions, **x(t)**, **y(t)** and **z(t)**, is equivalent to specifying a single vector function **r(t)** of the scalar independent variable **'t'**.Equation **(1.1)** is known as the \(\displaystyle\textbf{law of motion}\).Thus the law of motion **(1.1)** **specifies the position of the particle at any moment in time.**Therefore, determining the motion of a particle quantitatively means finding its law of motion ,i.e, equation **(1.1)**.

The velocity vector \(\displaystyle v={ v }_{ x }(t)+{ v }_{ y }(t)+{ v }_{ z }(t)\) and acceleration vector \(\displaystyle a={ a }_{ x }(t)+{ a }_{ y }(t)+{ a }_{ z }(t)\) are defined as the appropriate derivative of equation (1.1).Thus:

\[v=\cfrac { dr }{ dt } =\cfrac { dx }{ dt } \hat { i } +\cfrac { dy }{ dt } \hat { j } +\cfrac { dz }{ dt } \hat { k }\quad ...(1.2)\]

where

\[{ v }_{ x }(t)=\cfrac { dx }{ dt } ,\quad { v }_{ y }(t)=\cfrac { dy }{ dt } ,\quad { v }_{ z }(t)=\cfrac { dz }{ dt } \]

And

\[a=\cfrac { dv }{ dt } =\cfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } \hat { i } +\cfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } \hat { j } +\cfrac { { d }^{ 2 }z }{ d{ t }^{ 2 } } \hat { k }\quad ...(1.3)\]

where

\[{ a }_{ x }(t)=\cfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } ,\quad { a }_{ y }(t)=\cfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } ,\quad { a }_{ z }(t)=\cfrac { { d }^{ 2 }z }{ d{ t }^{ 2 } } \]

Equation **(1.1)** is fundamental to kinematics.Knowing the law of motion,one can determine any parameter of motion. What are parameters of motion? They are the physical quantities characterizing the motion.Mainly acceleration and velocity.

Broadly speaking, problems in kinematics can be classified as direct and indirect[Actually all sorts of problems be it physical or mathematical or... can be classified as direct and indirect.]The direct problem of Kinematics consists in finding any physical quantity characterizing the motion of the particle from the law of motion.In such problems, the law of motion is known or is given in the question itself.It is solved by consistently using the basic laws of kinematics **(1.1)- (1.3)**

The indirect problem of kinematics consists in determining the law of motion from other parameters of motion(velocity acceleration,etc.).That means anything but equation **(1.1)** is given and from what is given or known we have to determine equation **(1.1)**.

[Attempt only if you are quite experienced with Calculus and Vectors ]

Can you derive the following equations by concepts discussed here?Suppose the acceleration of the particle is **'a'** and remains constant.The velocity at time **0** be **'u'** and the velocity at time **'t'** be **' v'**.The particle is moving along the **x-axis.**

\[v=u+at\]

\[x=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\]

\[{ v }^{ 2 }={ u }^{ 2 }+2ax\]

Next post within 10/31/2014.Stay tuned.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestHm, what is precisely the

law of motion? I'm familiar withNewton's Laws of Motion, but not with any "law of motion" in particular. Anyway, above statements can be derived easily as follows:First, we know that velocity is the rate of change of position with respect to time; and, acceleration is the rate of change of velocity with respect to time. So, if we're given acceleration, all we need to ask is: "Derivative of what with respect to time will give me this [whatever "this" may be - that is given]?" This question is also known as the

antiderivative, or theINTEGRAL:\[\int{a(t)dt}=v(t)\]

Now, if we're given that the acceleration is constant and is defined as \(a\), we can proceed:

\[\int{(a)}dt=at+C\]

Now \(C\) here is actually the

initial velocity, commonly expressed as \(v_0\). The initial velocity is simply the velocity before the object's motion is taken into account - or, at \(t=0\) (and "before taken into account" simply implies that, for example, if we observe a bell being thrown up with some sort of velocity, we only care for the velocity after it is released, and not for any zig-zag arm movements along the ballbefore) - and hence \(v_0\).And so, if you plug in \(t=0\), you see \(C=v_0\) - or, in this case, \(u\). Thus,

\[v(t)=at+u\]

or,

\[\boxed{v=u+at}\]

Same reasoning applies next, except now it's initial

positioninstead of velocity. And so,\[\int{\left( u+at \right)}dt=ut+\frac{1}{2}at^2+C\]

Now here the assumption may be that for our observation purposes the initial position \(=0\), so we discount \(C\). But the general equation will have a shape of, where \(p(t)\) is the position function in terms of time:

\[p(t)=\frac{1}{2}at^2+v_0t+p_0\]

For our little derivation issue, we can express the position for our axis (hence \(x\)) as follows:

\[\boxed{x=ut+\frac{1}{2}at^2}\]

Now \(v^2\) can be defined in several ways, but, deriving from our past equations, it will go like this:

\[v^2=(u+at)^2=u^2+2aut+a^2t^2=u^2+2a(ut+\frac{1}{2}at^2)=u^2+2ax\]

\[\boxed{v^2=u^2+2ax}\]

Ta-da!

Once you've done this, try these out:

\[v=v_0+at \dots \dots \dots \dots \dots \boxed{x-x_0}\]

\[x-x_0=v_0t+\frac{1}{2}at^2 \dots \dots \dots \boxed{v} \]

\[v^2=v_0^2+2a(x-x_0) \dots \dots \dots \boxed{t}\]

\[x-x_0=\frac{1}{2}(v_0+v)t \dots \dots \dots \boxed{a}\]

\[x-x_0=vt-\frac{1}{2}at^2 \dots \dots \dots \boxed{v_0}\]

Notice that on the left are the equations of motion for when acceleration is

CONSTANT, and on the right are the missing quantities. Can you derive each of these just given \(\frac{d}{dt}(a)=0\)? And can you explain, for bonus points,WHYif \(\frac{d}{dt}(a)\neq 0\) these equations won't work?Have fun!

Log in to reply

Nice solution.

@John Muradeli is still dominating Physics in brilliant. :D

That equation which tells the position of a particle with respect to time can be thought of as a function.It is definite in nature, the rules of the game remain same all the time.It is a law in the sense that particle,wherever they are, whatever they are, follow it or we can make them follow it.Mathematically speaking, that is the way to describe motion of a particle.Whatever, you will find that kind of thing in some old Physics books.

Can't match your prowess of explanations but still giving a try:If acceleration isn't constant,the equations are not valid[we cannot derive them using the assumption that acceleration is variable].The only way, I know, to handle a variable acceleration case is using work energy theorem, not the mentioned 5 equations, not even deferential equations, calculus helps if there is a function defined or we can establish a relation.Given 5 equations can be derived in the same way as you did.We also have to take account, here, the initial position,initial velocity,acceleration,deceleration.

Bye @john muradeli

Won't hear from me for a long time :) .

....Keep Teaching Physics To Brilliant.... :)

Log in to reply