Question

Find the velocity v(t) and position x(t) of a particle of mass m that is subject to a

frictional force (retarding force or drag) proportional to the square of the velocity. Draw graphs for velocity and position against time.

Answer 1

To find:

The velocity v(t) and position x(t) of a particle of mass m that is subject to a frictional force (retarding force or drag) proportional to the square of the velocity. Draw graphs for velocity and position against time.

Solution:

As per the question , the frictional force be:

$\therefore \: f = - k {v}^{2} \: \: \: \: \: \: ......(k \: is \: a \: constant)$

$\implies\: a= - \dfrac{k {v}^{2} }{m}$

Let's calculate the v - t relationship:

$\implies\: \dfrac{dv}{dt} = - \dfrac{k {v}^{2} }{m}$

$\implies\: \dfrac{dv}{ {v}^{2} } =( - \dfrac{k }{m} ) \: dt$

$\displaystyle \implies\: \int\dfrac{dv}{ {v}^{2} } =( - \dfrac{k }{m} ) \: \int dt$

$\displaystyle \implies\: - \dfrac{1}{ v } =( - \dfrac{k }{m} ) \: t + c$

$\displaystyle \implies\: \dfrac{1}{ v } = \dfrac{kt }{m} - c$

$\displaystyle \implies\: \dfrac{1}{ v } = \dfrac{kt - mc }{m}$

$\boxed{ \displaystyle \implies\: v = \dfrac{m}{kt - mc }}$

$\displaystyle \implies\: \dfrac{dx}{dt} = \dfrac{m}{kt - mc }$

$\displaystyle \implies\: dx = m \times \dfrac{dt}{kt - mc }$

$\displaystyle \implies\: \int dx = m \times \int\dfrac{dt}{kt - mc }$

$\displaystyle \implies\: x = m \times \frac{1}{k} \bigg \{ln(kt - mc) \bigg \} + c'$

$\boxed{\displaystyle \implies\: x = \frac{m}{k} \bigg \{ln(kt - mc) \bigg \} + c'}$

• k , c , c' are all constants.

• 1st graph is v(t) graph and 2nd is x(t) graph.

• Kindle observe the nature of the graph , and do not scale it.