Question

A particle moving on a straight line so that its distance $s$from a fixed point at any time $t$ is proportional to ${t^n}$. If $v$ be the velocity and \[a\]the acceleration at any time then $\dfrac{{nas}}{{n - 1}} = $
A) $v$
B) ${v^2}$
C) ${v^3}$
D) $2v$

Answer 1

Hint:Velocity and acceleration both describe motion, but there is an important difference between them. The velocity is the rate of change of position and acceleration is rate of change of velocity.

Complete step by step answer:
We know that, velocity is the rate of change of displacement with respect to time
$ \Rightarrow v = \dfrac{{ds}}{{dt}}$ . . . (1)
And acceleration is rate of change of velocity
$ \Rightarrow a = \dfrac{{dv}}{{dt}}$ . . . (2)
It is given that displacement $s$is directly proportional to time by the relation,
$s\alpha {t^n}$
We would add a constant to remove the proportionality sign
$ \Rightarrow s = \beta {t^n}$ . . . . . (3)
Where $\beta = $Constant
By putting the value of displacement in equation (1), we get
$v = \dfrac{d}{{dt}}(\beta {t^n})$
$v = n\beta {t^{n - 1}}$ . . . . (4) $\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
By putting the value of velocity from equation (4) in equation (2), we get
$a = \dfrac{d}{{dt}}(n\beta {t^n})$
$ \Rightarrow a = n \times (n - 1)\beta {t^{n - 2}}$ . . . . (5) $\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
Therefore, by putting the value of displacement and acceleration in $\dfrac{{nas}}{{(n - 1)}}$
We get,
$\dfrac{{nas}}{{(n - 1)}} = \dfrac{{n \times n(n - 1)\beta {t^{(n - 2)}} \times \beta {t^n}}}{{(n - 1)}}$
We can simplify it as
\[\dfrac{{nas}}{{(n - 1)}} = {n^2}{\beta ^2}{t^{n - 2 + n}}\]
$ = {n^2}{\beta ^2}{t^{2n - 2}}$
\[ = {n^2}{\beta ^2}{t^{2(n - 1)}}\]
Since, all the terms are squares, we can write them together and get
\[\dfrac{{nas}}{{(n - 1)}} = {\left( {n\beta {t^{(n - 1)}}} \right)^2}\]
$\dfrac{{nas}}{{(n - 1)}} = {v^2}$ $\left( {\because v = n\beta {t^{n - 1}}} \right)$
 Therefore, From the above explanation option (B) ${v^2}$is correct.

Note: Not knowing the basics and formulae of derivatives can lead to a problem while solving such questions. It is important that you study basic calculus before starting to solve the questions of kinematics. Due to the lack of knowledge about derivatives, if you try to attempt it using the laws of motion, the answer will be incorrect as no information is given about initial velocity of the particle.