Question

The distance x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation, t = √x +3, where x is in metres and t in seconds. Find the displacement of the particle when its velocity is zero.

Answer 1

Given :

$\bullet\ \; \sf t=\sqrt{x}+3$

where ,

• t denotes time in seconds
• x denotes displacement in meters

To Find :

Displacement of the particle when velocity is zero

Knowledge Required :

We simply say that , displacement is zero because whenever velocity is zero , displacement must be zero

Instantaneous velocity is given by ,

$\bullet\ \; \sf v=\dfrac{dx}{dt}$

Solution :

$\sf t=\sqrt{x}+3\\\\:\implies \sf \sqrt{x}=t-3\\\\:\implies \sf x=(t-3)^2\\\\:\implies \sf x=t^2-6t+9\ \; \bigstar$

Let's find velocity ,

$:\implies \sf v=\dfrac{dx}{dt}\\\\:\implies \sf v=\dfrac{d}{dt}(t^2-6t+9)\\\\:\implies \sf v=2t-6\ \; \bigstar$

We need to find the displacement of the particle when velocity is zero ,

$:\implies \sf v=0\\\\:\implies \sf 2t-6=0\\\\:\implies \sf 2t=6\\\\:\implies \sf t=3\ s$

Let's find displacement when t = 3 s ,

$:\implies \sf x_{t=3\ s}=(3)^2-6(3)+9\\\\:\implies \sf x=9-18+9\\\\:\implies \sf x=18-18\\\\:\implies \sf \pink{x=0\ m}\ \; \bigstar$