Distance and velocity increased by $g$ units every second 

*

*Distance fallen in every second gets increased by $g$ units.

*Velocity is increased by $g$ units every second.


Is it true that distance and velocity increases by $g$ after every second?
If it starts from rest, then distance in the first second will be, by $s=ut+1/2at^2$, $5 \;\text{m}$. Then if it increases by $g$, will it be $15 \;\text{m}$?
Similarly, the velocity, by $v=u+at$, will it be $10 \;\text{m/s}$ in the first second? Will it increase like $20 \;\text{m/s}$, $30 \;\text{m/s}$, $40 \;\text{m/s}$, ... every second?
P.S. $g$ is acceleration due to gravity, which I assume to be $10 \;\text{m}\,\text{s}^{-2}$.
 A: 
Is it true that distance and velocity increases by $g$ after every second?

That's true for velocity. It's not true for distance, but that's not what your quoted statement says.


*

*Distance fallen in every second gets increased by $g$ units.


That's talking about the additional distance covered in each second, not the total distance, which increases quadratically, that is, in proportion to $t^2$. Here's a table for the first 5 seconds, using $g = 10 \,\mathrm{m/s}$ and $s = \frac12 gt^2$




time
distance
increase




0
0



1
5
5


2
20
15


3
45
25


4
80
35


5
125
45




In each second, the amount of distance covered in that second is $g$ metres greater than the distance covered in the previous second.

FWIW, Galileo measured and described falling motion in this way, in terms of the additional distance covered per unit time.
A: An object falling has a constant acceleration $g$. This means at any time $t$ its velocity will be given by $$v=v_0+gt$$ If it has an initial velocity $v_0=0$ then $$v=gt$$ So its velocity will increase by a factor $10$ every second. In other words,
after one second it will have a velocity $10ms^{-1}$
and after two seconds it will have a velocity $20ms^{-1}$
and after three seconds it will have a velocity $30ms^{-1}$
and after $t$ seconds it will have a velocity $v=10\times t \ ms^{-1}$
So for every consecutive second the velocity has changed by $10ms^{-1}$.
We say it has an acceleration of $10$ meters per second, per second or "$10$ meters per second squared".
As you have stated, the distance it falls will follow a different relationship given by $$x=v_0t+\frac{1}{2}gt^2$$ or $$x=\frac{1}{2}gt^2$$ if $v_0=0$. Then
after one second it will have fallen $5m$
and after two seconds it will have fallen $20m$ change=$15m$ $\Delta$change=$10m$
after three seconds it will have fallen $45m$ and change$=25m$ $\Delta$change=$10m$
after four seconds it will have fallen $80m$ and change$=35m$ $\Delta$change=$10m$
after five seconds it will have fallen $125m$ and change$=45m$ $\Delta$change=$10m$
and after $t$ seconds it will have travelled $x=5\times t^2\ m$ and in this instance, the difference between the distances after consecutive seconds will always be $10m$.
There is a quadratic relationship between $x$ and $t$ as oppose to the linear relationship between $v$ and $t$, but differences shown above after each second are always 10 units. So both of the statements


*

*Distance fallen in every second gets increased by $g$ units.

*Velocity is increased by $g$ units every second.


are true but the first statement talks about additional distance after each second.
