Question

An aircraft carrier weighting 6 times 10^7 kg is traveling at 20 m/s on the open ocean when it runs out of fuel. It is then slowed by a fluid frictional force given by F = - beta e^alpha v where beta = 3 times 10^5 N and alpha = 0.1 s/m. Using Newton's Second Law, F = mv', find when the aircraft carrier will come to a stop by solving the initial value problem then setting v = 0. Notice that you need not solve explicitly for the velocity in order to be able to answer this question. However, you'll want to solve explicitly for the velocity, anyway, as you will use this result in part (b). Report your answer in seconds to 2 decimal places.

Answer 1

Fluid Frictional Force acting on aircraft is $F= \beta e^{\alpha v}$

Now by newton's laws of motion,

Comparing above two equations, ( as we neglected air friction)

$mdv/dt= \beta e^{\alpha v}$

$dv/e^{\alpha v}= (\beta/m) dt$

Integrating above equation, we get

   $ln(e^{\alpha v})/ (ve^{\alpha v})= (\beta/m)t + C$ ( where C= constant)

   $\alpha v/ (ve^{\alpha v})= (\beta/m)t + C$   ( because: )

   $\alpha/e^{\alpha v}= (\beta /m)t+ C$ ................(1)

Now given, at t=0s, v= 20 m/s

Using above values in equation,

$\alpha/e^{20 \alpha }= 0+ C$

or $C= \alpha/e^{20 \alpha }$

Putting above in equation1 ,

$\alpha/e^{\alpha v}= (\beta /m)t + \alpha/e^{20 \alpha }$

$\alpha(1/e^{\alpha v}-1/e^{20 \alpha }) = (\beta /m)t$

Now when v= 0, then value of time is

$\alpha(1/e^{\alpha 0}-1/e^{20 \alpha }) = (\beta /m)t$

$\alpha(1/1-1/e^{20 \alpha }) = (\beta /m)t$

Using given values, $(0.1)(1/1-1/e^{20(0.1)) }) = ((3 \times 10^5) /(6 \times 10^7))t$

   $(0.1)(1-1/ 7.389}) = (5 \times 10^{-3})t$

$(0.1)(1-0.135) = (5 \times 10^{-3})t$

( ANS)