Physics, asked by darkmatterT01, 8 months ago

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.

Answers

Answered by nirman95
3

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.

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