Question

Q8. Deduce the position-velocity relation using the velocity-time graph.​

Answer 1

Equation for position velocity relation consider graph given in figure.

We know that distance travelled 's' by a body in time 't' is given by the area under line AB.

which is area of trapezium OABC .

So we have

Distance travelled = s = Area of Trapezium OABC

S =(sum of parallel sides) × height/2

=(OA+CB) x OC/2

since, OA + CB = u+v and OC = t

Then we get,

S = (u+v) /2 __ ( 1 )

from velocity time relation,

t= v-u/a

substituting the value of 't' in ( 1 )

we get,

S=(u+v) / 2(v-u) /2

or we have,

v²=u²+2ac

which is equation for position velocity relation.

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