Question

Please answer

You are swimming underwater in a large lake when you find a rock and decide to throw it down (all while fully immersed in the water). If one assumes that the relationship between fluid resistance and speed is given by this equation: f-kv, then compute the speed of the rock as a function of time as it travels down in the water given that you throw it with a speed of 3mg/k. (Note that k is the coefficient in the above fluid resistance equation.) Express your answer in terms of vr (terminal speed), k, m, and t.

Answer 1

We know that the terminal speed of the rock will be when net force will be Zero.
Therefore
f- mg = 0
kv = mg
at this point velocity would be terminal therefore , v = v_T
kv_T = mg
v_T = mg/k ------------(1)
Now at any point
mg - kv = ma
we know that acceleration is given by
a= dv/dt , therfore
mg - kv = m*(dv/dt)
Now ,
$\left ( \frac{m}{mg-kv} \right )dv = dt$
Now integrating both side , we get
$\int \left ( \frac{m}{mg-kv} \right )dv =\int dt$
$\left ( \frac{-m}{k} \right )ln\left ( mg-kv \right ) = t+C$
where C is integration constant
$ln\left ( mg-kv \right ) =\left ( t+C \right )\left ( \frac{-m}{k} \right )$
taking anti-log both side
$\left ( mg-kv \right ) =e^{\left ( t+C \right )\left ( \frac{-m}{k} \right )}$
$kv =mg - e^{\left ( t+C \right )\left ( \frac{-m}{k} \right )}$
Now at t = 0 , v = 3mg/K
$k\left ( \frac{3mg}{k} \right ) =mg - e^{\left ( 0+C \right )\left ( \frac{-m}{k} \right )}$
$3mg =mg - e^{\left ( C \right )\left ( \frac{-m}{k} \right )}$
$2mg = e^{\left ( \frac{k}{mC} \right )}$
on taking log both side
$ln2mg = {\left ( \frac{k}{mC} \right )}$
$C = {\left ( \frac{k}{m*ln(2mg)} \right )}$
Therefore the equation of v will be
$kv =mg - e^{\left ( t+C \right )\left ( \frac{-m}{k} \right )}$
$v =\frac{mg}{k} - \left ( \frac{1}{k} \right )e^{\left ( t+C \right )\left ( \frac{-m}{k} \right )}$
using the equation of v_T
$v =v_{\tau } - \left ( \frac{1}{k} \right )e^{-\left ( \frac{mt}{k} + \frac{1}{ln(2mg)}\right )}$