Question

The relationship between time $t$ and displacement $x$ is $t=\alpha x^{2}+\beta x$, where $\alpha$ and $\beta$ are constant, find the relation between velocity and acceleration.

Answer 1

We know that the rate of change of displacement is called velocity and the rate of change of velocity is called acceleration. Here, instead a value for displacement and time, an equation is given. So calculate velocity and acceleration, we need to differentiate the equation.

Formulas used:
$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$
Also,$\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}$

Complete step by step answer:
To find the relationship between $v$ and $a$, let us start with the basic definition of velocity and acceleration.
We know that the velocity $v$ i.e. $v=\dfrac{displacement}{time}$. Or $v=\dfrac{dx}{dt}$ where time is $t$ and displacement is $x$.
Similarly, the acceleration $a$ is defined as the rate of change of velocity i.e. $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$ where, time is $t$ and velocity $v$ .
Since, $t$is given in terms of $x$, we can differentiate $t$ with respect to $x$, then we get $\dfrac{dt}{dx}=2\alpha x+\beta$
Then velocity $v=\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}=\dfrac{1}{2\alpha x+\beta}$
Then, we must differentiate $v=\dfrac{dx}{dt}=\dfrac{1}{2\alpha x+\beta}$ with respect $t$. Here, using the mathematical differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, here, in our sum, $n=-1$. And using chain rule of differentiation, we get
$a=\dfrac{d^{2}x}{dt^{2}}= -1(2\alpha x+\beta)^{-2}\times 2\alpha =\dfrac{-2\alpha}{(2\alpha x+\beta)^{2}}$
To find the relationship between $v$ and $a$, we can substitute $v=\dfrac{1}{2\alpha x+\beta}$ in $a$
Then we get, $a=-2\alpha v^{2}$
Hence the relationship between $v$ and $a$, is$a=-2\alpha v^{2}$

Note:
This may seem as a hard question at first. But this question is easy, provided you know differentiation, here we use the mathematical differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, here, in our sum, $n=-1$. And using chain rule of differentiation, we get the result.
Also see that $\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}$, this is the most important step in this question. Also note that $v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$. To calculate, $a$ we must differentiate only $v$ with respect to $t$ and not $\dfrac{dt}{dx}$.