Question

A constant force acts on a particle and its displacement x (in cm) is related to the time t (in s) by the equation t=sqrtx +3, when the velocity of the particle is zero, its displacement ( in cm) is

Answer 1

Given,

A constant force acts on a particle and its displacement varies with time as,

$\: \: \: \: \: \: \: \: \: t = \sqrt{x} + 3$

Velocity is given by,

$\: \: \: \: \: \: \: \: \: v = \frac{dx}{dt}$

Differentiating the given equation with respect to t gives,

$\implies \frac{d}{dt} (t) = \frac{d}{dt} ( \sqrt{x } + 3) \\ \\ \implies \: 1 = \frac{1}{2 \sqrt{x} } \frac{dx}{dt} + 0 \\ \\ \implies \: 1 = \frac{1}{2 \sqrt{x} } \times v \\ \\ \implies \: v = 2 \sqrt{x}$

So, The relation between velocity and displacement is given by,

$v = 2 \sqrt{x}$

When the velocity is zero,

$0 = 2 \sqrt{x} \\ \\ 0 = \sqrt{x} \\ \\ x = 0$

Therefore, The displacement is zero when the velocity is zero

Answer 2

Answer:

Given:

Considering the Time Vs Displacement relationship to be as follows :

$t = \sqrt{(x + 3)}$

To calculate:

Displacement of particle when Velocity is zero.

Concept:

We will try to differentiate the given function and get the Displacement vs To function :

$t = \sqrt{(x + 3)}$

$= > x + 3 = {t}^{2}$

$= > \dfrac{d(x + 3)}{dt} = \dfrac{d {t}^{2} }{dt}$

$= > \dfrac{dx}{dt} + 0 = 2t$

$= > v = 2t$

Now , let's find the time for which velocity is zero.

Putting v = 0 , we get t = 0 ;

Now putting t = 0 in the 1st equation , we get :

$t = \sqrt{(x + 3)}$

$= > 0 = \sqrt{(x + 3)}$

$= > x + 3 = 0$

$= > x = - 3 \: cm$

So final answer :

$\boxed{ \large{ \blue{ \bold{ displacement = - 3 \: cm}}}}$