Question

A boat of mass m was travelling with speed u when the boatman stopped rowing.
Given the retarding force due to water to be ae^bv at speed v (a and b are constants),
find the time required by the boat to stop. Assuming the water to be stationary

Answer 1

Given:

A boat of mass m was travelling with speed u when the boatman stopped rowing.

Given the retarding force due to water to be ae^bv at speed v

To find:

Time taken by boat to stop ?

Calculation:

$\therefore \: force = a {e}^{bv}$

$= > \: acceleration = f= \dfrac{a {e}^{bv} }{m}$

$= > \: f= \dfrac{a {e}^{bv} }{m}$

$= > \: \dfrac{dv}{dt} = \dfrac{a {e}^{bv} }{m}$

$= > \: \dfrac{dv}{ {e}^{bv} } = \dfrac{a }{m} \: dt$

$= > \: {e}^{ - bv} \: dv = \dfrac{a }{m} \: dt$

Integrating on both sides:

$= > \: \displaystyle \int {e}^{ - bv} \: dv = \dfrac{a }{m} \: \int dt$

Putting limits:

$= > \: \displaystyle \int_{u}^{0} {e}^{ - bv} \: dv = \dfrac{a }{m} \: \int_{0}^{t} dt$

$= > \: \displaystyle \dfrac{ - 1}{b} \bigg( {e}^{0} - {e}^{ - bu} \bigg)= \dfrac{a }{m} \: \bigg(t - 0 \bigg)$

$= > \: \displaystyle \dfrac{ - 1}{b} \bigg( 1 - {e}^{ - bu} \bigg)= \dfrac{at }{m}$

$= > \: \displaystyle \dfrac{ 1}{b} \bigg( {e}^{ - bu} - 1 \bigg)= \dfrac{at }{m}$

$= > \:t = \displaystyle \dfrac{ m}{ab} \bigg( {e}^{ - bu} - 1 \bigg)$

So, final answer is:

$\boxed{ \bold{ \:t = \dfrac{ m}{ab} \bigg( {e}^{ - bu} - 1 \bigg)}}$