Derive graphically the equation of motion which relate position and velocity
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The First Equation of Motion. Consider an object moving so that its velocity-time graph is a straight line. Such an object is undergoing constant acceleration, since the slope of thegraph is constant. The first equation of motion gives the final velocity after a time t for these objects, given an initialvelocity v0 .
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s=Area of ∆ABC + Area of rectangle OADC
s=1/2×Base×Height + Length×Breadth
s=1/2×AC×BC + OA×AC
s=1/2×t×(v-u) + t×u............(I)
Now,v-u=at
v-u/a=t
Put the value of 't' in equation(I),we have
s=1/2×(v-u)×v-u/a+u×[(v-u/a)]
s=(v-u)^2×2u×(v-u)
___________
2a
s=v^2+u^2-2uv+2uv-2u^2
___________________
2a
s=v^2-u^2
_______
2a
2as=v^2-u^
s=1/2×Base×Height + Length×Breadth
s=1/2×AC×BC + OA×AC
s=1/2×t×(v-u) + t×u............(I)
Now,v-u=at
v-u/a=t
Put the value of 't' in equation(I),we have
s=1/2×(v-u)×v-u/a+u×[(v-u/a)]
s=(v-u)^2×2u×(v-u)
___________
2a
s=v^2+u^2-2uv+2uv-2u^2
___________________
2a
s=v^2-u^2
_______
2a
2as=v^2-u^
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