Derive a position velocity relatrelation
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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) x height/2
=(OA+CB)x OC/2
since OA+CB=u+v and OC=t,
Then we gt
s=(u+v)t/2get,-------------(1)
from, velocity time relation,
t=v-u/a
substituting the value of 't' in equation 1
we get,
s=(u+v)/2(v-u)/2
or we have
v²=u²+2as
which is 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) x height/2
=(OA+CB)x OC/2
since OA+CB=u+v and OC=t,
Then we gt
s=(u+v)t/2get,-------------(1)
from, velocity time relation,
t=v-u/a
substituting the value of 't' in equation 1
we get,
s=(u+v)/2(v-u)/2
or we have
v²=u²+2as
which is equation for position velocity relation
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Answer:
Airport elevation is the reduced level above mean sea level of high level of transport
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