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  1. Distance fallen in every second gets increased by $g$ units.
  2. 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 $10 \;\text{m}\,\text{s}^{-2}$.

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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.

  1. 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.

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  • $\begingroup$ A table on an SE site... I never knew that was possible. $\endgroup$
    – Steeven
    Sep 9 at 11:27
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    $\begingroup$ @Steeven We got Markdown tables late last year. See meta.stackexchange.com/q/356997/334566 Of course, on Physics.SE we can also use MathJax tables. $\endgroup$
    – PM 2Ring
    Sep 9 at 11:30
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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

  1. Distance fallen in every second gets increased by $g$ units.
  2. Velocity is increased by $g$ units every second.

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

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