Physics Help Forum A body starts from origin with initial velocity U and retardation KV^3...?

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 Aug 28th 2008, 08:22 AM #1 Junior Member   Join Date: Aug 2008 Posts: 10 A body starts from origin with initial velocity U and retardation KV^3...? A body starts from origin with initial velocity U and retardation KV^3 where V is instantaneous velocity of the body. A) velocity of the body at any instant is √(2U^2Kt + 1) B) velocity of the body at any instant is U/[√(2U^2Kt + 1)] C) the x coordinates of the body at any instant is 1/UK * [√(2U^2Kt + 1) - 1] D) Magnitude of initial acceleration of the body is KU^3 More than one options may be correct. __________________ 先生
Aug 29th 2008, 06:35 AM   #2
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 Originally Posted by sensei A body starts from origin with initial velocity U and retardation KV^3 where V is instantaneous velocity of the body. A) velocity of the body at any instant is √(2U^2Kt + 1) B) velocity of the body at any instant is U/[√(2U^2Kt + 1)] C) the x coordinates of the body at any instant is 1/UK * [√(2U^2Kt + 1) - 1] D) Magnitude of initial acceleration of the body is KU^3 More than one options may be correct.
$\displaystyle a = -kv^3$

(which means that option D is correct)

$\displaystyle \Rightarrow \frac{dv}{dt} = -kv^3 \Rightarrow \frac{dt}{dv} = -\frac{1}{kv^3}$ where v = U when t = 0

$\displaystyle \Rightarrow t = \frac{1}{2kv^2} + C$.

Substitute v = U when t = 0: $\displaystyle C = -\frac{1}{2kU^2}$

$\displaystyle \Rightarrow t = \frac{1}{2kv^2} - \frac{1}{2kU^2} = \frac{1}{2k} \left( \frac{1}{v^2} - \frac{1}{U^2} \right)$.

Solve for v as a function of t. You get option B.

To check option C you could differentiate it and see if you get option B ....

 Tags body, initial, kv3, origin, retardation, starts, velocity
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