The position of the particle moving in one dimension, under the action of constant force is given by equation

$straight t equals square root of straight x plus 3$

where x is in meter and t is in second. Find the displacement of the particle when its velocity is momentarly zero.

Solution

We have,
Position of the particle moving under a constant force, $straight t equals square root of straight x plus 3$
$rightwards double arrow$ $square root of straight x equals straight t minus 3$
$rightwards double arrow$ x = t²-6t + 9
Differentiating w.r.t. r, we get
$straight v space equals space dx over dt equals straight d over dt left parenthesis straight t squared minus 6 straight t plus 9 right parenthesis$

$rightwards double arrow$ v = 2t - 6, is the velocity of the particle.
The time at which the velocity of the particle will be zero is given by,

$space space space space space space 2 straight t minus 6 equals 0 space rightwards double arrow space space straight t space equals space 3 straight s$
Now,
Initial position of the particle is,
$straight x subscript 0 equals left parenthesis 0 right parenthesis squared minus 6 left parenthesis 0 right parenthesis plus 9 equals 9$
Position of the particle at t = 3 is given by,
x=(3)²- 6(3) + 9 = 0
$space space therefore$ Displacement of the particle is,
x - x₀= 0 - 9
= - 9m