# Free Fall and constant acceleration

Although the acceleration of free fall is constant, why don't the distance go like $$y = 9.8+4.9= 14.7m$$ after 2 seconds, $$y= 19.6+14.7 = 34.3m$$ after 3 seconds? I think Constant acceleration work like say, start accelerating with $$5ms^{-2}$$, so the velocity is increasing $$5ms^{-1}$$ every second and the distance is like $$5m,15m, 30m\dots$$So second question is my thought on constant acceleration is right?

I think Constant acceleration work like say, start accelerating with $$5ms^{−2}$$, so the velocity is increasing $$5ms^{−1}$$ every second and the distance is like $$5m,15m,30m\dots$$So second question is my thought on constant acceleration is right?

It's wrong and the cause of your misconception. It's only the velocity of an object increasing constantly over time when constant acceleration occurs not the distance. A body which is moving with constant acceleration has also increase in its velocity within a second. So, it doesn't have a constant change in displacement over time. Also the distance covered by a free falling object is calculated by the equation $$S = \frac {1}{2}gt²$$

Where $$S$$ is the displacement or distance covered by a free falling body, $$g$$ is gravitational acceleration and $$t$$ is time.

Hope this helps.

If the acceleration is constant at say $$5\,\rm m\,s^{-2}$$ and say at one instant of time the velocity is $$10\,\rm m\,s^{-1}$$.
At the end of one second the velocity will be $$15\,\rm m\,s^{-1}$$ and during that one second interval the velocity is continuously changing.
Because the acceleration is constant, the distance covered during that one second interval is the average velocity ($$(10+15)/2=7.5 \,\rm m\,s^{-1}) \,\times \,$$ the time interval ($$1\,\rm s$$) $$= 7.5\,\rm m$$.

• Average velocity × time = $ut + \frac {1}{2}at²$ Jun 1 at 8:43

It's not that simple.

Speed in calculus is defined as $$v = \frac {dx}{dt}$$,

so

$$dx = v dt$$,

integrating both sides we get :

$$x = \int v dt ~~~~~~~~~~~~[1]$$

However speed in this case is not constant, namely $$v=v(t)=at$$,

so substituting into equation (1), we get :

$$x = \int at dt$$

And $$a$$ (acceleration) is constant in this case (though in general case it may be not), so it can be put before integral (it's integration rule,- all constants can be factored out of integral), so :

$$x = a \int tdt ~~~~~~~~~~~~~~~[2]$$,

And it's a known math fact that $$x^n$$ has antiderivative of $$\large {\frac {x^{n+1}}{n+1}}$$, when $$n \ne -1$$,

so integral at (2) reduces to :

$$x = a \int t^1dt = a \frac {t^{1+1}}{1+1} = a \frac {t^2}{2}$$.

### Conclusion

You need to learn basic calculus if you want to understand why concrete mathematical/physics function goes one way or the other. That's why Newton+Leibniz has invented it for solving physics problems.