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.