please explain the question fully
A particle starts from rest with uniform acceleration a its velocity after n second is V the displacement of the body in the last 2 seconds is
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we know the formula distance travelled in nth second=u+a/2(2n-1)
=a/2(2n-1)
and so the distance travelled in (n-1)the second=a/2{2(n-1)-1}
=a/2(2n-2-1)
=a/2(2n-3)
so the distance travelled in last two seconds=distance travelled in last second+distance travelled in second last second
=a/2(2n-1)+a/2(2n-3)
=a/2{2n-1+2n-3)
=a/2(4n-4)
=a/2×2(2n-2)
=a(2n-2)
answer can be expressed in other variables also plzz say me the options
=a/2(2n-1)
and so the distance travelled in (n-1)the second=a/2{2(n-1)-1}
=a/2(2n-2-1)
=a/2(2n-3)
so the distance travelled in last two seconds=distance travelled in last second+distance travelled in second last second
=a/2(2n-1)+a/2(2n-3)
=a/2{2n-1+2n-3)
=a/2(4n-4)
=a/2×2(2n-2)
=a(2n-2)
answer can be expressed in other variables also plzz say me the options
sanidhyasingla:
we have to tell the displacement of the body from the starting to the last second
ok we have to find the displacement of the body in only last 2 seconds
i think you should do displacement in last second - displacement in second last second
to find displacement in last 2 second
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