Question

A body is moving according to the equation $x=at+bt^{2}-ct^{3}$, where $x$ is the displacement, $a\;,b\;and\;c$ are constants. The acceleration of the body is

& A.a+2bt \\\ & B.2b+6ct \\\ & C.2b-6ct \\\ & D.3b-6c{{t}^{2}} \\\ \end{aligned}$$

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 acceleration, we need to differentiate the equation to find the velocity and differentiate the velocity to obtain the acceleration.

Formula:
$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$

Complete answer:
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 there is a $a$ term in the equation, let $\;acc$ denote the acceleration of the equation.
Here, it is given that,$x=at+bt^{2}-ct^{3}$, clearly, $x$ is given in term of $t$.Here, using the mathematical differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, then velocity $v=\dfrac{dx}{dt}=a+2bt-3ct^{2}$
Then, to find the acceleration $acc=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$we must differentiate$v=\dfrac{dx}{dt}=a+2bt-3ct^{2}$ with respect$t$. Using $\dfrac{d}{dx}x^{n}=nx^{n-1}$, again, then we get, $acc=2b-6ct$
Since we know that also, $\dfrac{d}{dx}K=0$ where $K$ is a constant and is independent of $x$, thus the $a$ term vanishes in the acceleration equation.
Hence the answer is $a=2b-6ct$

Therefore, the correct option is C.

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 $\dfrac{dt}{dx}$.