# How acceleration affects velocity?

I understood that the acceleration changes the velocity and the velocity changes the position.

So I tried to calculate the position of a falling object, where $$y_{acc} = 9.81$$ and the initial values for position and velocity are: \begin{align} y &= 0\\ y_{vel} &= 0 \end{align}

So, In my understanding on each second $$y_{acc}$$ is added to $$y_{vel}$$ after that $$y_{vel}$$ is added to $$y$$. Taking this approach I get this values for $$y(t)$$.

\begin{align} y(0) &= 0 \\ y(1) &= 9.81 \\ y(2) &= 29.43 \\ y(3) &= 58.86 \\ y(4) &= 98.1 \\ y(5) &= 147.15 \\ \end{align}

But these results differ from this formula $$y(t) = \frac{1}{2}at^2$$, where I got this values for $$y(t)$$

\begin{align} y(0) &= 0 \\ y(1) &= 4.905 \\ y(2) &= 19.62 \\ y(3) &= 44.145\\ y(4) &= 78.48 \\ y(5) &= 122.625 \end{align}

I would like to know what did I got wrong; Aditional note: I'm trying to do a simulation, but I do not want to rely on a $$\Delta t$$.

• Your steps should be infinitesimally small in place of one second. Commented Jul 28 at 8:43
• Have you heard of FDTD? Commented Jul 28 at 8:57
• @my2cts yeah... I think that's correct... yikes idk what to do, and nope I have not head of FDTD Commented Jul 28 at 9:04
• In which grade are you? Commented Jul 28 at 9:52
• If you plot your results you will see how to set your step size and how to correct to get a more accurate result for finite steps. Commented Jul 28 at 9:58

... each second $$y_{acc}$$ is added to $$y_{vel}$$ ...

Not really. $$y_{vel}$$ changes continuously, not just once a second. The overall effect is to increase $$y_{vel}$$ by an amount $$y_{acc}$$ each second, but this only works because $$y_{acc}$$ is constant.

... after that $$y_{vel}$$ is added to $$y$$ ...

No - otherwise $$y$$ would be incremented in jerky steps, once every second. $$y$$ changes continuously, and the amount by which it changes is determined by $$y_{vel}$$ at each instant - and $$y_{vel}$$ is also changing continuously.

Computer simulations cannot update quantities continuously (otherwise it would take an infinite number of updates and hence an infinite time to simulate any finite interval). However, you can approximate the true equation of motion $$y = \frac 1 2 a t^2$$ by updating $$y$$ and $$y_{vel}$$ once per time step $$\Delta t$$. If you take $$\Delta t$$ as $$1$$ second you get (as you found) a not very accurate approximation. If you use a smaller value for $$\Delta t$$ then you get a better (but still not perfect) approximation.

Simulating the motion of objects in this way is called numerical integration. Numerical integration with a constant $$\Delta t$$ timestep is just one method - there are many other methods of numerical integration, some of which are more accurate than others.

• I tried to make the aceleration smaller (dividing it), and it seems that the result is closer to the $\frac{1}{2}at^2$ resutls. Commented Jul 28 at 17:05

You probably receive more down votes for bad notation, but good job for trying. Anyway, the problem is here:

So, In my understanding on each second 𝑦𝑎𝑐𝑐 is added to 𝑦𝑣𝑒𝑙 after that 𝑦𝑣𝑒𝑙 is added to 𝑦

For a falling object, and in absence of air resistance, we can assume that acceleration is constant, but of course velocity is not constant. What does that imply?

Well, acceleration is the rate of change of velocity. That means, for our case of constant acceleration, in every second the object's velocity increases by a constant amount.

On the other hand, velocity is the rate of change of position. If velocity was constant, we could say that position would change by a constant amount in every second. But we know that is not the case.

Let us consider the first 1 second of the free fall. At the beginning of the fall, velocity is 0. After one second, velocity is 9.81 m/s. Certainly it is wrong to assume that during that second, the object's velocity was 0 all along, and suddenly became 9.81 m/s. Likewise, it is wrong to assume that the velocity was 9.81 m/s all during that time.

If you would like to build a computer model, here is a trick you can use, only for this particular example:

In each time interval (for example from t=0s to t=1s), you can assume that the velocity of the falling object is equal to the velocity of the middle time of that interval (in the above example: t=0.5s).

Good luck

• yeah... I'm not familiar with the notation; so I can not sum the velocity to the position if the velocity is not constant? Commented Jul 28 at 7:30
• @clararaquel In general, no! But in this particular example you can use average velocities. For example: $y(1) = y(0) + v(0.5).1$ Commented Jul 28 at 7:51

You error is perhaps best illustrated using a velocity vs time graph.

You have stated quite correctly that the velocity after 1 second is $$g$$,
after 2 seconds is $$2g$$, after 3 seconds is $$3g$$, etc.

So the displacement,
So the displacement,
after 1 second is $$g\,({\rm metre/second}) \times 1\,({\rm second})=g$$ metres,
after 2 seconds is $$g \times 1+2g \times 1=3g$$ metres,
after 3 seconds is $$g \times 1+2g \times 1+3g \times 1=6g$$ metres. You have worked out the sum of the green shaded rectangles.

What you have not accounted for is the fact that the velocity changes during the one second interval and so you should be using the average velocity in each of the time intervals to work out the displacement.

So the correct displacement,
after 1 second is $$g/2\times 1=g/2$$ metres,
after 2 seconds is $$g/2 \times 1+3g/2 \times 1=2g$$ metres,
after 3 seconds is $$g/2 \times 1+3g/2 \times 1+5g/2 \times 1=9g/2$$ metres.

However as the graph is a straight line you can choose any time interval.
Let's choose a time interval $$t$$ and so, starting form rest, the velocity after time $$t$$ is $$gt$$ and the average velocity in that time interval is $$gt/2$$.
Hence the displacement is $$gt/2 \times t = \frac 12 gt^2$$.

• Okay I think I see a pattern, every second we add a term of the form $x\frac{g}{2} \times 1$ but I don't get what is $x$?, It is 1, 3, 5, and after? Commented Jul 28 at 9:32
• It is the multiplication sign! $3 \times 2 = 6$ To get the displacement one need to multiply the velocity $v$ by the time interval, $\Delta t$, $v \times \Delta t$. Commented Jul 28 at 10:49