Answer to Question #165204 in Mechanics | Relativity for Dora

Question

The relation (3t=3x+6) describes the displacement of a particle in one direction. Where x is in metres and t in seconds. Find the displacement when the velocity is zero

Answer 1

The relation $3t=\sqrt{3}x+6$  describes the displacement of particle in one direction where x is in meter and t is in second.

We have to find displacement when velocity is zero. i.e. $\dfrac{dx}{dt}$

Now, $3t=\sqrt{3}t+6$

$3t-6=\sqrt{3}x$

$(3t-6)^2=3x$

$x = \dfrac{9t^2-36t+36}{3}$

Differentiate w.r.t 't'

$\dfrac{dx}{dt} = \dfrac{18t-36}{3} = 0$

$t=2s$

Hence, the displacement at $t = 2$ is

$x = 0$ m