EQUATIONS OF MOTION

Let us consider an object moving with initial velocity (u), after time (t), it attains final velocity (v), covering distance (s) with uniform acceleration (a). 

 First equation of motion-

VELOCITY-TIME RELATION (v=u+at)


OA= u 
BC= v 
BD= v-u 
OC= t
                  Acceleration= Change in velocity ÷
                                            Time taken 
        a= BC-DC / OC = BD/OC 
        a= v-u/t (cross multiplication)  
    u+at =v (arrange it according to equation)

Hence, first equation v=u+at is proved. 

Second equation of motion 

POSITION-TIME RELATION (s=ut+½ at²)


 S= Area under the graph
S= Area of rect.(OADC)+Area of triangle ABD
               S=OA×OC+½ AD×BD 
               S= ut+½(t) (v-u)  
Put the values of (v-u) from eq 1 [v=u+at]
                                                            [v-u=at]
s=ut+½(t)(at) (arrange according to eq)

Hence, second equation s=ut+½at² is proved. 

Third equation of motion 

POSITION-VELOCITY RELATION (2as=v²-u²)


    S= Area under the graph 
    S= Area of trapezium OABDC
    S= ½ (Sum of all parallel sides) × Height 
    S= ½ (BC+OA) × OC
    S= ½ (v+u) (t)
Put the value of the from eq 1 [v=u+at]
                                                        [v-u=at]
                                                        [v-u/a=t]
S= ½(v+u)(v-u/a)
S= 1/2a (v+u) (v-u) [a+b][a-b]= a²-b²
2as =(v+u)(v-u) 

Hence, third equation 2as=v²-u² is proved. 



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