Deriving the three eq:of motion with its graph
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distance travelled= area of trapezium
distance(s)=sum of parallel side×height÷2
s=(OA+CB)×OC÷2
now,oa+cb=u+v and oc=t
s=u+v×t÷2
v=u+at
at=v-u
t=v-u÷a
s=u+v×v-u÷2a
because v+u×v-u=vsquare-u square
2as=vsquare-usquare
v2=u2+2as
distance(s)=sum of parallel side×height÷2
s=(OA+CB)×OC÷2
now,oa+cb=u+v and oc=t
s=u+v×t÷2
v=u+at
at=v-u
t=v-u÷a
s=u+v×v-u÷2a
because v+u×v-u=vsquare-u square
2as=vsquare-usquare
v2=u2+2as
alpanagottam:
Wrong
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