Question

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

Answer 1

⭕️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

$\huge\bf{\boxed {\red{\mathfrak{thank \: you :)}}}}$

Answer 2

Given information :

Initial velocity ( u ) = 0

[ the body started from rest ]

Final velocity ( v ) = v₁

time interval ( t ) = t₁

mass = m.

Formulas to use :

v = u + at

F = ma

W = F . S

Finding the answer

v = u + at

==> v = 0 + at

==> v = at

==> a = v/t

Put the values of t and v

==> a = v₁/t₁

Now :

F = ma

  ==> mv₁/t₁

Find the displacement

S = ut + 1/2 at²

= 1/2 at² [ since S = 0 ]

S = 1/2 v₁/t₁ × (t)²

==> S = 1/2 v₁/t₁ × t²

W = F × S

==> mv₁/t₁ + 1/2 v₁/t₁ × t²

==> 1/2 m v₁² / t₁² × t²

Power = W / t

==> [ 1/2 m v₁² / t₁² × t² ] / t

==> 1/2 m v₁² t  / t₁

ANSWER :

P = 1/2 m v₁² t  / t₁

Hope the answer is correct!

Hope it helps !

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