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 X = t^2 -6t + 9, where x is in meter and t is in seconds. find the displacement of the particle when its velocity is 0

Answer 1

differentiating x and equating the value with 0 would give the time and putting time in the value of x would give the displacement.

Answer 2

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.