Question

A particle is moving along $x - axis$. If the position of the particle varies with time as $x \propto {t^n}$ (where $n$ is any constant), then select the correct option ($t \ne 0$)
A. If $n > 2$, then acceleration of particle is increase with time
B. If $n < 0$, the acceleration of particle is decrease with time
C. If $n > 2$, the velocity of particle is uniform
D. Both (1) and (2)

Answer 1

In the given question we have to use first the single derivative of distance with respect to time which gives velocity, then we have to check the equation whether it is increasing or not. The double derivative then gives the acceleration, which upon the given options we have to check whether it follows aforesaid statements or not.

Complete step by step answer:
The particle is moving in the $x - axis$. In the given question, it is there that,
$x \propto {t^n} - - - - - \left( 1 \right)$
Derivative of equation $\left( 1 \right)$ with respect to $t$ we get,
$\dfrac{{dx}}{{dt}} \propto n{t^{n - 1}} - - - - - \left( 2 \right)$
We know $\dfrac{{dx}}{{dt}}$ is known as velocity. Let us now check the statements about velocity given in the options. When $n > 2$, the velocity of particle is,
Let us take $n = 3$ we get the equation as, $\dfrac{{dx}}{{dt}} \propto 3{t^2}$
Again, $n = 4$, $\dfrac{{dx}}{{dt}} \propto 4{t^3}$
Thus, when $n > 2$ the velocity of the particle keeps on increasing rapidly with time. Hence it is non-uniform.So, option (C) is incorrect.

Again, derivative of equation $\left( 2 \right)$ with respect to $t$ we get,
$\dfrac{{{d^2}x}}{{d{t^2}}} \propto n\left( {n - 1} \right){t^{n - 2}} - - - - - \left( 3 \right)$
$\dfrac{{{d^2}x}}{{d{t^2}}}$ is known as acceleration.
Let us now check the statements mentioned in the options, when $n > 2$ for equation $\left( 3 \right)$ we get a positive quantity of the answer.
Let us take $n = 3$, we get $\dfrac{{{d^2}x}}{{d{t^2}}} \propto 6t$
Again when $n = 4$, we get, $\dfrac{{{d^2}x}}{{d{t^2}}} \propto 12{t^2}$
Thus it keeps on increasing with increasing time.So, option $\left( A \right)$ is correct.

When $n < 0$ the value t is always negative. Let us check for $n = - 1$, we get $\dfrac{{{d^2}x}}{{d{t^2}}} \propto 2{t^{ - 3}}$
$n = - 2$ then, $\dfrac{{{d^2}x}}{{d{t^2}}} \propto 6{t^{ - 4}}$
Hence, it is clear that it keeps on decreasing with time. So, option (B) is also correct.Now, we conclude that option (A) and (B) are both correct.

Hence, the correct answer is option D.

Note: The first derivative of distance with time represents velocity while the second derivative represents acceleration. The velocity which does not change with time is known as uniform velocity. The graph of a uniform velocity is a straight line whereas uniform acceleration is a constant slope.