Answer:
See below
Step-by-step explanation:
Write the initial value problem and the solution for the downward velocity for an object that is dropped (not thrown) from a great height.
if v(t) is the speed at time t after being dropped, v'(t) is the acceleration at time t, so the the initial value problem for the downward velocity is
v'(t) = 10 - 0.1v(t)
v(0) = 0 (since the object is dropped)
The equation v'(t)+0.1v(t)=10 is an ordinary first order differential equation with an integrating factor
[tex]\bf e^{\int {0.1dt}}=e^{0.1t}[/tex]
so its general solution is
[tex]\bf v(t)=Ce^{-0.1t}+100[/tex]
To find C, we use the initial value v(0)=0, so C=-100
and the solution of the initial value problem is
[tex]\bf \boxed{v(t)=-100e^{-0.1t}+100}[/tex]
what is the terminal velocity?
The terminal velocity is
[tex]\bf \lim_{t \to\infty}(-100e^{-0.1t}+100)=100\;mt/sec[/tex]
How long before the object reaches 90% of terminal velocity?
90% of terminal velocity = 90 m/sec
we look for a t such that
[tex]\bf -100e^{-0.1t}+100=90\rightarrow -100e^{-0.1t}=-10\rightarrow e^{-0.1t}=0.1\\-0.1t=ln(0.1)\rightarrow t=\frac{ln(0.1)}{-0.1}=23.026\;sec[/tex]
How far has it fallen by that time?
The distance traveled after t seconds is given by
[tex]\bf \int_{0}^{t}v(t)dt[/tex]
So, the distance traveled after 23.026 seconds is
[tex]\bf \int_{0}^{23.026}(-100e^{-0.1t}+100)dt=-100\int_{0}^{23.026}e^{-0.1t}dt+100\int_{0}^{23.026}dt=\\-100(-e^{-0.1*23.026}/0.1+1/0.1)+100*23.026=1,402.6\;mt[/tex]
