Question

Acceleration -kv^3 find final velocity

Answer 1

AnsweR :

• $\sf{v \: = \: \dfrac{1}{\sqrt{2kt \: - \: 2c}} \: ms^{-1}}$

ExplanatioN :

$\longrightarrow \sf{a \: = \: - \: kv^3 \: \: \: \: \: ...(1)}$

Also, the formula for instantaneous acceleration is :

$\longrightarrow \sf{a \: = \: \dfrac{dv}{dt} \: \: \: \: \: ...(2)}$

Now, equate (1) and (2)

$\longrightarrow \sf{\dfrac{dv}{dt} \: = \: -kv^3} \\ \\ \longrightarrow \sf{\dfrac{dv}{v^{-3}} \: = \: -kdt} \\ \\ \longrightarrow \sf{dv^{-2} \: = \: -kdt}$

Now, integrate both the sides :

$\longrightarrow \sf{\int dv^{-2} \: = \: \int -kt} \\ \\ \longrightarrow \sf{\dfrac{v^{-2}}{-2} \: = \: -kt \: + \: c} \\ \\ \longrightarrow \sf{v^{-2} \: = \: -2(-kt \: + \: c)} \\ \\ \longrightarrow \sf{v^{-2} \: = \: 2kt \: - \: 2c} \\ \\ \longrightarrow \sf{v \: = \: \dfrac{1}{\sqrt{2kt \: - \: 2c}}} \\ \\ \boxed{\sf{v \: = \: \dfrac{1}{\sqrt{2kt \: - \: 2c}}} \: ms^{-1}}$

Answer 2

Answer:

Given:

Acceleration is given as follows :

$a = - k {v}^{ 3}$

To find:

Relationship between velocity and time

Concept:

Acceleration refers to instantaneous change of velocity with respect to time. Hence it can be written in differential form as follows :

$a = \dfrac{dv}{dt}$

During integration, since enough information hasn't been provided , we will show only the indefinite Integral.

Calculation:

$\bigstar \: \: - k {v}^{ 3} = \dfrac{dv}{dt}$

$= > \dfrac{dv}{ {v}^{ 3} } = - k \: dt$

$= > {v}^{-3} \: dv = - k \: dt$

Integrating on both sides :

$\displaystyle = > \int{v}^{-3} \: dv = \int - k \: dt$

$= > \dfrac{ {v}^{-2} }{-2} = - kt + c$

$= > {v}^{-2} = 2kt + c$

$= > \dfrac{1}{v} = \sqrt{ (2kt+c)}$

c is a constant .