Question

The motion of a body falling from rest in a resisting medium described by the equation a= A-BV ..where a and b are constants then find the velocity at any instant t

Answer 1

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Answer 2

Answer:

a and b are constants then the velocity at any instant t is $v=\frac{A}{B}\left(1-e^{-B t}\right)$

Explanation:

Velocity exists essentially as a vector quantity. It is the rate of change distance. It exists the rate of change of displacement. The velocity of an object moving can never be negative. The velocity of a moving object can be zero.

Here,

$\frac{d v}{d t}=(A-B v)$

$\int_{0}^{v} \frac{d v}{(A-B v)}=\int_{0}^{t} d t$

$-\left.\frac{1}{B} \log _{e}(A-B v)\right|_{0} ^{v}=\left.t\right|_{0} ^{t}$

Then,

$-\frac{1}{B}\left[\log _{e}(A-B v)-\log _{e} A\right]=\left.t\right|_{0}(t)$

$\log _{e}\left(\frac{A-B v}{A}\right)=-B t$

$(A-B v)=A e^{-B t}$

Therefore,

$B v=A-A e^{-B t}$

$v=\frac{A}{B}\left(1-e^{-B t}\right)$

So, a and b are constants then the velocity at any instant t is $v=\frac{A}{B}\left(1-e^{-B t}\right)$.

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