Question

A body of mass 10 slugs is dropped from a height of 1000ft with no initial velocity. The body encounters an air resistance proportional to its velocity. If the limiting velocity is known to be 320ft/s, find an expression for the velocity of the body at any time t, an expression for the position of the body at any time t, and the time required for the body to attain a velocity of 160ft/s.

Answer 1

\[F_g+F_{air}=ma\]Since the air resistence is proportional to velocity and opposing gravity, we have:\[-bv+mg=ma\]

Answer 2

\[1 = \frac{ma}{mg-bv}\]
\[\int 1dt=\int \frac{ma}{mg-bv}dt = \int\frac{m}{mg-bv}\frac{dv}{dt}dt = \int\frac{m}{mg-bv}dv \]

Answer 3

The left hand integral is easy.\[\int 1dt=t+c\]The right hand integral is a bit tricky.\[\int \frac{m}{mg-bv}dv=m \int \frac{1}{mg-bv}dv\]We're going to have to do \(u\) substitution for this.

Answer 4

\(u=mg-bv\) and \(du=-bdv \rightarrow \frac{du}{-b}=dv\)
So we have:
\[m \int \frac{1}{u} \frac{du}{-b}= \frac{-m}{b}\ln(u)+c_2\]Then we undo the substitution to get:\[\frac{-m}{b}\ln(mg-bv)+c_2\]And going back to our equation:
\[t+c = \frac{-m}{b}\ln(mg-bv)+c_2\]

Answer 5

Let \(k=c-c_2\), they're just constants anyway
\[t+k = \frac{-m}{b}\ln(mg-bv)\]\[(t+k)\frac{-b}{m} = \ln(mg-bv)\]To get rid of the \(\ln\) we have to raise both sides as a power of \(e\). Also let \(k_2 = \frac{-bk}{m}\) since it's just a constant.
\[\Large e^{\frac{-b}{m}t+k_2}=mg-bv\]\[\Large k_3e^{\frac{-bt}{m}}=mg-bv\]

Answer 6

\[ \Large k_3e^{\frac{-bt}{m}}=mg-bv\]\[ \Large k_3e^{\frac{-bt}{m}}+bv=mg\]\[ \Large bv = mg- k_3e^{\frac{-bt}{m}}\]\[ \Large v = \frac{mg}{b} - k_4e^{\frac{-bt}{m}}\]Notive that this is a function with respect to time.

Answer 7

\[ \Large v(t) = \frac{mg}{b} - k_4e^{\frac{-bt}{m}}\]

Answer 8

To keep things general, let's assume that we didn't know initial velocity, and use the variable \(v_i\) for it....\[\Large v(0)= v_i\rightarrow v_i = \frac{mg}{b} - k_4e^{\frac{-b(0)}{m}}\]\[v_i=\frac{mg}{b}-k_4e^0= \frac{mg}{b}-k_4\]\[v_i+k_4=\frac{mg}{b}\]\[k_4=\frac{mg}{b}-v_i\]

Answer 9

So our general equation for velocity which has proportional air resistance is:
\[\Large v(t) = \frac{mg}{b} -(\frac{mg}{b}-v_i)e^{\frac{-bt}{m}}\]

Answer 10

We just have to use the information given in the problem to solve for \(b\), since we already have \(m, g, v_i\).

Answer 11

To solve for \(b\), given some terminal velocity \(v_t\), we recall that at terminal velocity, gravity and air resistance are equal. Thus: \[-bv_t-mg=0\rightarrow b=\frac{-mg}{v_t}\]

Answer 12

What is g?