body of mass m accelerates uniformly from rest to velocity v1 in time interval T1 . the instantaneous power delivered to the body as a funcyion of time it is.
answer should be= mv12 t/T12
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Body of mass m accelerates uniformly from rest to velocity v1 in time interval T1 . the instantaneous power delivered to the body as a funcyion of time it is.
we know that :-
⇒ v = u + at
now , let's take
⇒ u = 0
⇒ t = T1
⇒ a = v÷T1
so, force
⇒ F = ma = m(v÷T1)
∴
⇒ s = ut + (1÷2)at²
⇒since u = 0
⇒∴ s = (1÷2)at² = (1÷2) (v÷T1)×t²
and work done will be
W = F.s = m(v÷T1) ×(1÷2) (v÷T1)×t²
W = (1÷2) m (v÷T1)²×t²
∴ finally the instantaneous power will be====
P = W/t = [(1÷2).m.(v÷T1)²×t²] / t
⇒∴ P = (1÷2).m.(v÷T1)²×t
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ANSWER
Let a be the constant acceleration
Using 1st equation of motion
v1=u+aT1
v1=0+aT1
a=T1v1..........(i)
At any instant t, the velocity v of the body
v=0+at
v=T1v1⋅t.........(ii)
∵P=Fv
P=mav
P=m(T1v1)(T1v1⋅t)
P=T12mv12t
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