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(C) Given the velocity of the particle is $\frac{dx}{dt} = \cos^2(\pi x)$.To find the time $t$ taken to reach a position $x$,we separate the variables

Vedclass · Accepted Answer

$x = \frac{1}{2}$

Vedclass · Answer

(C) Given the velocity of the particle is $\frac{dx}{dt} = \cos^2(\pi x)$.To find the time $t$ taken to reach a position $x$,we separate the variables:$dt = \frac{dx}{\cos^2(\pi x)} = \sec^2(\pi x) dx$.Integrating both sides from the origin $(x=0, t=0)$ to a position $x$ at time $t$:$\int_0^t dt = \int_0^x \sec^2(\pi u) du$.$t = \left[ \frac{	an(\pi u)}{\pi} ight]_0^x = \frac{	an(\pi x)}{\pi}$.As $x 	o \frac{1}{2}$,$	an(\pi x) 	o 	an(\frac{\pi}{2}) 	o \infty$.Therefore,$t 	o \infty$ as $x 	o \frac{1}{2}$.This implies that the particle takes an infinite amount of time to reach the point $x = \frac{1}{2}$,meaning it never reaches this point.

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

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