The first derivative of displacement is velocity.
The second derivative of displacement is acceleration (which also means acceleration is the first derivative of velocity)

So, to show that the particle is decelerating, means to show that the acceleration is negative.


i.e

d²s/dt² < 0

s = ln(2t + 3)

ds/dt = 2/(2t + 3)

d²s/dt² = (-1)(2/(2t + 3)¹+¹)(2)

= -4 / (2t + 3)²


Since t ≥ 0 , (time cannot be negative)

2t + 3 ≥ 3 for all real non-negative values of t

Which means that 2t + 3 > 0

So (2t + 3)² also > 0
(The square of a positive value is still positive)

Then, -4 / (2t + 3) < 0 for all real non-negative values of t.
(Dividing a negative value -4 by an always positive value (2t + 3) will always yield a negative value)

does displacement always have to be the distance from the start point to another point or it can start from any point?

You might want to read up on the definition of displacement.


https://en.m.wikipedia.org/wiki/Displacement_(geometry)


Basically you have 1 starting point.

The displacement is the shortest change in position from that point to the end point (i.e a straight line)

It can be positive or negative.