Dear Indrayani

    v= u + at   
    s =( u + v ) t /2    
    s = ut + at2/2 
    v2 = u2 + 2as    

The following are the variables used in these equations:

    t is the length of the interval of time and measured in seconds.
    u is the initial speed of the object or the speed at the beginning of the time interval, measured in ms-1.
    v is the final speed of the object or the speed at the end of the time interval, measured in ms-1.
    a is the acceleration of the object within the time interval t measured in ms-2.
    s is the distance covered or the displacement of the object during the time interval t, measured in metres.

Derivation of Newton's Equations of Motion:
Derivation of First and second equations of motion:
We know that,
Velocity is the Change in displacement / Change in time,
v = ds/dt = s/t   and
Acceleration is the change in velocity / change in time,
a = dv/dt = (v-u)/t
Or the acceleration is the change in speed per unit time, so:
a = (v-u) /t
or at = v-u
or v= u + at    and it is the first Equation of Motion.

We also know that the average speed during the time interval t is equal to displacement per unit time, so:
(u + v)/2 =  s / t   since Average speed = (u+v)/2
s / t = (u + v)/2
or   s = ( u + v ) t /2 and this is the second equation of motion.

Derivation of third and Fourth Equations of Motion:
Derivation of third equation of motion
The third equation of motion can be derived by substituting the value of v from the first equation into the second equation.
First equation is v= u +at
And the second equation is s = (u+v)t /2    
Or  s=[u + (u+at)] t /2   =[2u+at]t /2 = [2ut+ at2] / 2 =ut + at2/2
Or  s=ut + at2/2  and this is  the third equation of motion.

Derivation of fourth equation of motion
To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it.
Acceleration a = (v-u) /t
Or at = (v-u)
Or t = (v-u) /a

Next step is to substitute the equation for the time t into the second equation of motion s=ut + at2/2   to obtain:
s = u[(v-u)/a]  + a [(v-u)/a]2 /2 
or s = u(v-u)/a  + (v-u)2 /2a
or s =[2u+v-u](v-u)/2a
or  s = (u+v)(v-u)/2a = (v+u)(v-u)/2a
or  s= (v2-u2)/2a 
or  2as = v2 –u2
or v2 –u2  =   2as    
or v2 = u2 + 2as     and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

All the best.
AKASH GOYAL
AskiitiansExpert-IIT Delhi

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively: Click here to download the toolbar.
						

							S=(u+v)t/2Derivation-We know that a=(v-u)/t and s=ut+(at^2)/2S=ut+((v-u)/t)t^2S=ut+(v-u)t/2Take t as commonS=t[2u+v-u]/2S=t(v+u)/2 Proved
						

							
we know that ( u + v )/2 is the average velocity in case of constant accelaration.
we also know that average velocty into time is total displacement.
now we have (u+v)/2*t=average velocity * t
which can also be written as (u+v)t/2=s