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# An object with mass $m$ is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If $s(t)$ is the distance dropped after $t$ seconds, then the speed is $v = s'(t)$ and the acceleration is $a = v'(t).$ If $g$ is the acceleration due to gravity, them the downward force on the object is $mg - cv,$ where $c$ is a positive constant, and Newton's Second Law gives$m \frac {dv}{dt} = mg - cv$(a) Solve this as a linear equation to show that $v = \frac {mg}{c} (1 - e^{-ct/m})$ (b) What is the limiting velocity?(c) Find the distance the object has fallen after $t$ seconds.

## a)$$v(t)=(m g / c)\left[1-e^{-(c / m) t}\right]$$b) $$\lim _{t \rightarrow \infty} v(t)=m g / c$$c)$$(m g / c)\left[t+(m / c) e^{-(c / m) t}\right]-m^{2} g / c^{2}$$

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Differential Equations

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