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I'm wondering how I can get the final velocity of a particle given its acceleration, displacement and initial velocity.

I know that there's a formula to get it but I am trying to figure out the result intuitively.

  • A freely falling object dropped from an airplane,that is going downards at a velocity of $10m/s^2$(gravity), what is its velocity is at $50m$.

Getting the result by the formula I get

$v_f^2 = v_i+2a \Delta x$ $\ \ \ $ = $v_f=\sqrt{0^2+10\cdot2\cdot50} = 31.62277$

and now I want to get the same result by intuition

The acceleration is $10m/s^2$, at the first second the velocity is $10m/s$ and the distance travelled is $10m$, after the second second the velocity is $20m/s$, and the distance travelled is $30m$, after the third second the velocity is $30m/s$, and the distance travelled is $60m/s$. But wait, since the velocity at $50m$ is $31.62277m/s$ by the formula, why does at $60m$ the velocity is $30m/s$ ?.

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  • $\begingroup$ In your first equation you've said that it is $50$ seconds when in the question it is the distance that is $50m$. Is this a mistake or no? $\endgroup$ – CooperCape Nov 15 '17 at 17:46
  • $\begingroup$ @CooperCape It is $\Delta x$, already fixed it. $\endgroup$ – hjx Nov 15 '17 at 17:48
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    $\begingroup$ Regarding your reasoning in the final paragraph; the fact that the velocity at the end of one second is 10 m/s doesn't imply that the distance travelled is 10 m. In fact, the distance travelled is 5 m. Can you tell me why this is? $\endgroup$ – Alfred Centauri Nov 15 '17 at 17:52
  • $\begingroup$ @AlfredCentauri why doesn't 10m/s imply that the distance travelled is 10m ? $\endgroup$ – hjx Nov 15 '17 at 18:19
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    $\begingroup$ hjx, if the average velocity is a 10 m/s over the first second then you can say that the distance traveled in that second is 10 m. But the average velocity cannot be 10 m/s since the velocity is zero initially and just reaches 10 m/s at the end of 1 second. $\endgroup$ – Alfred Centauri Nov 15 '17 at 19:15
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Firstly, you made a fundamental error in assuming that the distance traveled after $1\text{s}$ would be $10\text{m}$ if accelerating at $10\text{m s}^{-2}$. To realize why this is an error, you need to remember that the particle's velocity isn't a constant $10\text{m s}^{-1}$ for the whole $1\text{s}$; rather, the velocity is increasing from $0\text{m s}^{-1}$ to $10\text{m s}^{-1}$ at a constant rate.

Velocity vs. Time graph for the first second of acceleration: enter image description here

As you can see from the above graph, the velocity of the particle increases linearly, and the average velocity over the second can be seen (intuitively) to be $5\text{m s}^{-1}$. This shows that the distance traveled by the particle over the first second is $5\text{m}$. Similarly, for the second second, the particle will accelerate uniformly between $10\text{M s}^{-1}$ and $20\text{m s}^{-1}$, yielding an average velocity over the interval of $15\text{m s}^{-1}$, which causes a distance traveled of $15\text{m}$.

It is much easier to calculate this using the formula you named, but if you want to do this by intervals, you can use the above demonstrated method to find your answer sequentially.

Finding velocity after $50$ meters sequentially:

$$5\text{m} \Rightarrow 10\text{m s}^{-1}$$ $$5\text{m} + 15\text{m} = 20\text{m}\Rightarrow 20\text{m s}^{-1}$$ $$5\text{m} + 15\text{m} + 25\text{m} = 45\text{m} \Rightarrow 30\text{m s}^{-1}$$ $$\color{red}{5\text{m} + 15\text{m} + 25\text{m} + 35\text{m} = 80\text{m} \Rightarrow 40\text{m s}^{-1}}$$

Now, you might be able to see why there may be problem with following this "intuitive" process to find the velocity... How do you move from $45\text{m}$ to $50\text{m}$? Well, you could guess and say that it's something just over $30\text{m s}^{-1}$, and that is really as far as you can get without delving into kinematic equations.

The math to solve for the velocity at $50\text{m}$ from here is not really that complicated, but if you're going to solve it algebraically, might as well just apply the formula $v_2^2=v_1^2+2a\Delta x$ in the first place.

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  • $\begingroup$ Wonderful answer, thank you. But please clear up another thing, why does the average velocity is equal to the distance traveled ? $\endgroup$ – hjx Nov 15 '17 at 23:19
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    $\begingroup$ The distance traveled is not "equal" to the average velocity—it is equal to the average velocity multiplied by the time elapsed. If you were to be measuring in intervals of two seconds, the distance traveled would be two times the average velocity over the interval. $\endgroup$ – Kieran Moynihan Nov 15 '17 at 23:33
  • $\begingroup$ Desmos Graphing Calculator for image? $\endgroup$ – Alfred Centauri Nov 16 '17 at 0:32
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Just focus on the first second. As you said at the end of the first second the velocity is $10 m/s$. If something travels at $10 m/s$ for a whole second, it will have moved 10m. But the object doesn't travel that speed for the entire first second. That's just how fast its traveling at the END of the second.It increases during the entire second and at the very end of one second it has just barely reached a velocity of $10 m/s$ The object spends almost the entire first second traveling less than $10 m/s$. So if an object that travel's $10 m/s$ for a whole second will travel 10m, how can an object that travels less than 10$m/s$ for the first $.99999999999$ seconds travel the same distance?

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(For what it's worth now that someone else has posted a visual aide, here's what I had drawn up earlier for an intuition builder ...)

enter image description here

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