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 is v 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).

#### Solution:

Show me the final answer↓

Remember that we can write acceleration as:

a(v)=\dfrac{dv}{dt}

dt=\dfrac{dv}{a(v)}

dt=\dfrac{dv}{a(v)}

Let us take the integral of both sides of the equation:

\,\displaystyle \int^{t_2}_{t_1}dt=\int^{v}_{v_0}\dfrac{1}{a(v)}dv

Substitute our acceleration equation:

\,\displaystyle \int^{t}_{0}dt=\int^{v}_{0}\dfrac{1}{9.81[1-v^2(10^{-4})]}dv

t=\dfrac{50}{9.81}\text{ln}\left(\dfrac{1+0.01v}{1-0.01v}\right)

Isolate for v: v=\dfrac{100(e^{0.1962t}-1)}{e^{0.1962t}+1}

t=\dfrac{50}{9.81}\text{ln}\left(\dfrac{1+0.01v}{1-0.01v}\right)

Isolate for v: v=\dfrac{100(e^{0.1962t}-1)}{e^{0.1962t}+1}

At t=5 seconds, we have:

v=\dfrac{100(e^{(0.1962)(5)}-1)}{e^{(0.1962)(5)}+1}

v=45.5 m/s

v=45.5 m/s

As t\rightarrow\infty, we have:

v=100 m/s

#### Final Answers:

v=45.5 m/s

Maximum velocity, v=100 m/s

Maximum velocity, v=100 m/s

further explanation as to how v = 100 m/s for maximum velocity would be appreciated!

As the value of t increases, the values in the brackets (so the “e” values) will cancel out with the bottom, leaving just 100.