Question:
The solution of $\frac{dv}{dt} +\frac{k}{m}v = -g$ is
The solution of $\frac{dv}{dt} +\frac{k}{m}v = -g$ is
Updated On: Apr 19, 2024
- $v = ce^{-^{\frac{k}{m}t}} - \frac{mg}{k}$
- $v = c- \frac{mg}{k} e^{-^{\frac{k}{m}t}}$
- $v e^{-^{\frac{k}{m}t}} = c- \frac{mg}{k}$
- $v e^{^{\frac{k}{m}t}} = c- \frac{mg}{k}$
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The Correct Option is A
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$\frac{dv}{dt}+\frac{k}{m}v = -g \Rightarrow \frac{dv}{dt} = -\frac{k}{m}\left(v+\frac{mg}{k}\right)$
$\Rightarrow \frac{dv}{v+mg/k} = -\frac{k}{m}dt \Rightarrow log \left(v+\frac{mg}{k}\right)$
$= -\frac{k}{m}t+log\,C$
$\Rightarrow v+\frac{mg}{k}= Ce^{-kt/m} \Rightarrow v = Ce^{-kt/m} -\frac{mg}{k}$
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The correct option is A)
\(\frac{dt}{dv}+\frac{k}{m}v=-g\)
⟹\(\frac{dv}{dt}=\frac{k}{m}(v+\frac{mg}{k})\)
⟹ \(\frac{dv}{v+\frac{mg}{k}}=\frac{k}{m}dt\)
⟹ \(log(v+\frac{mg}{k})=-\frac{k}{m}t+log\,c\)
⟹ \(v+\frac{mg}{k}=c^{-(\frac{k}{m})\times t}\)
⟹ \(v=ce^{(\frac{k}{m})t}-\frac{mg}{k}\)
\(\frac{dt}{dv}+\frac{k}{m}v=-g\)
⟹\(\frac{dv}{dt}=\frac{k}{m}(v+\frac{mg}{k})\)
⟹ \(\frac{dv}{v+\frac{mg}{k}}=\frac{k}{m}dt\)
⟹ \(log(v+\frac{mg}{k})=-\frac{k}{m}t+log\,c\)
⟹ \(v+\frac{mg}{k}=c^{-(\frac{k}{m})\times t}\)
⟹ \(v=ce^{(\frac{k}{m})t}-\frac{mg}{k}\)
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.