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If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81[1 - { v }^{ 2 }({ 10 }^{ -4 })] m/{ s }^{ 2 }, where v is in m/s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s , and (b) the body’s terminal or maximum attainable velocity (as t \rightarrow \infty).

Step-by-step

Velocity: The velocity of the particle can be related to the time.

\begin{aligned}(+ \downarrow)\quad\quad & dt = \frac { dv } { a } \\ & \int_{ 0 }^{ t } { dt } = \int_{ 0 }^{ v } { \frac { dv } { 9.81[1 – (0.01v)^{ 2 }] } } \\ & t = \frac { 1 } { 9.81 }[\int_{ 0 }^{ v } { \frac{ dv } { 2(1 + 0.01v) } + \int_{ 0 }^{ v } { \frac { dv } { 2(1 – 0.01v) } } }] \\ & 9.81t = 50\text{ ln }(\frac { 1 + 0.01v } { 1 – 0.01v }) \\ & v = \frac { 100({ e }^{ 0.1962t } – 1) } {{ e }^{ 0.1962t } + 1} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \text{(1)} \end{aligned}

a) When t = 5 s , then, from Eq. (1)

v = \frac { 100[{ e }^{ 0.1962(5) } – 1] } { { e }^{ 0.1962(5) }\space\space +\space\space 1 } = 45.5 \text{m/s}

b) If t \rightarrow \infty, \frac { { e }^{ 0.1962t } – 1 } { { e }^{ 0.1962t }\space\space +\space\space 1 } \rightarrow 1. Then, from Eq. (1)

{ v }_{ max } = 100 \text{ m/s }

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