Question

The displacement x of a paraticle moving in one dimension under the action of a constant force is related to time by the equation t=√x+3.Find the displacement if particle when its velocity =0​

Answer 1

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

Given Equation,

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

$x = (t-3)^2\\\\x = t^2 - 6t + 9$

Velocity = dx / dt.

Diffrentiate the equation w.r.t t:

$\dfrac{dx}{dt}=\dfrac{d(t^2 - 6t + 9)}{dt} \\\\v = 2t - 6\\\\0 = 2t - 6\\2t = 6\\t = 3$

When t = 3,:

$t = \sqrt x + 3\\\\3 = \sqrt x + 3\\\\x = 0$

Hence Displacement is zero.

Answer 2

Given,

The particle is under constantly accelerated motion.

Relation between t and x, $t=\sqrt{x}+3$

v = 0

To find,

The displacement of the particle when its velocity, v=0

Solution,

This can easily be solved using the equation of velocity with displacement in differential form,

$v=\frac{dx}{dt}$    _____ (1)

$t=\sqrt{x}+3$       (according to the question)

⇒$\sqrt{x} = t-3$

⇒$x = (t-3)^{2}$

⇒$x = t^{2} - 6t + 9$ _____(2)

from equation (1) and (2) we can derive,

v = $\frac{d(t^{2}-6t+9) }{dt}$

⇒v= 2t - 6

∵ v = 0

⇒ 2t-6 = 0

⇒t = 3  _____(3)

From equation (2) and (3),

x = 0, (when v = 0)

∴ Displacement 'x' of the particle when v = 0, is 0.