Question

the relation between time t and distance X is t= ax²+ßx , where a and ß are constants. The retardation is ?​

Answer 1

★ Concept :-

Here the concept of differentiation has been used. We see that we are given a relationship and thus we have to find the retardation of the body. So firstly we can differentiate the given equation with respect to time to find the value of velocity. Then again we can differentiate it with respect to time to find the acceleration.

Let's do it !!

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★ Solution :-

*Note : Here ß = b .

Given,

$\;\bf{\mapsto\;\;\green{t\;=\;ax^{2}\;+\;bx}}$

Now let's differentiate this equation with respect to time (t) first.

$\;\bf{\rightarrow\;\;\dfrac{dt}{dt}\;=\;\dfrac{dx}{dt}ax^{2}\;+\;\dfrac{dx}{dt}bx}$

$\;\bf{\rightarrow\;\;\dfrac{dt}{dt}\;=\;\dfrac{dx}{dt}2ax\;+\;\dfrac{dx}{dt}bx}$

$\;\bf{\rightarrow\;\;\dfrac{dt}{dt}\;=\;2ax\dfrac{dx}{dt}\;+\;b\dfrac{dx}{dt}}$

Since a and b are constants and x is raised to 2.

We know that,

→ v = dx/dt

where v is the velocity. So,

$\;\bf{\rightarrow\;\;\dfrac{dt}{dt}\;=\;2axv\;+\;bv}$

Now taking v as common we get,

$\;\bf{\rightarrow\;\;\dfrac{dt}{dt}\;=\;v(2ax\;+\;b)}$

We know that,

→ dt/dt = 1

where t is the time. So,

$\;\bf{\rightarrow\;\;1\;=\;v(2ax\;+\;b)}$

$\;\bf{\rightarrow\;\;v(2ax\;+\;b)\;=\;1}$

$\;\bf{\rightarrow\;\;2ax\;+\;b\;=\;\dfrac{1}{v}}$

$\;\bf{\rightarrow\;\;\red{2ax\;+\;b\;=\;v^{-1}}}$

Now again differentiating this equation with respect to time, we get

$\;\bf{\rightarrow\;\;\dfrac{dx}{dt}(2ax\;+\;b)\;=\;\dfrac{dv}{dt}v^{-1}}$

Simce differentiation is distributive in nature,

$\;\bf{\rightarrow\;\;\dfrac{dx}{dt}2ax\;+\;\dfrac{dx}{dt}b\;=\;\dfrac{dv}{dt}(-v^{-2})}$

Since b is constant and x is to the power 1 , so

$\;\bf{\rightarrow\;\;2a\dfrac{dx}{dt}\;=\;(-v^{-2})\dfrac{dv}{dt}}$

We know that,

→ acceleration = dv/dt

acceleration = dv/dt→ velocity = dx/dt

By applying these in the equation, we get

$\;\bf{\rightarrow\;\;2av\;=\;(-v^{-2}) acceleration}$

Let the acceleration be r .

$\;\bf{\rightarrow\;\;2av\;=\;-v^{-2}\:\times\:r}$

$\;\bf{\rightarrow\;\;\dfrac{2av}{-\:v^{-2}}\;=\;r}$

$\;\bf{\rightarrow\;\;-\dfrac{2av}{v^{-2}}\;=\;r}$

$\;\bf{\rightarrow\;\;-2av^{3}\;=\;r}$

$\;\bf{\rightarrow\;\;r\;=\;-2av^{3}}$

Since acceleration = retardation.

This is the required answer.

$\;\underline{\boxed{\tt{Retardation\;=\;\bf{\purple{-2av^{3}}}}}}$