I seem to have misunderstood something in velocity. There is a path from point A to point B, that gap between is 100 meters. An object moves from point A to B with a velocity is 10 m/s.

$V=\frac{\Delta x}{\Delta t}$

$ x = 100$ $meters$

$ t = 10$ $seconds$

Now, within that gap is area x, is the velocity the same knowing that $x$ now is smaller than $100$$m$? The velocity would be less correct?

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Also, given the acceleration I still can find the velocity of the object correct? Does it matter what formula I should use?

Edit : The object has a constant acceleration that is: $1$ $m/s^2$

  • $\begingroup$ Why do you think the velocity would be less? $\endgroup$
    – David Z
    Commented Jun 9, 2014 at 6:53
  • $\begingroup$ Based on the formula above? If we want to know that velocity at that point that distance (x) would be less, and the time should be shorter? $\endgroup$
    – Pupil
    Commented Jun 9, 2014 at 7:09
  • $\begingroup$ If the acceleration is a constant 1m/s then the equation of motion is $s = 0.5at^2$ (where $a = 1$). So after 10 seconds the object would only have moved 50m not 100m. $\endgroup$ Commented Jun 9, 2014 at 7:48

1 Answer 1


If the motion is not accelerated, then yes, the velocity within the specified portion would remain the same. If the acceleration is uniform, then there are standard equations to use. If acceleration is non-uniform ( in most questions, the formula for a(t) is specified, like, a(t)=(t^2)+3t...as an instance), and varies wrt. time, you can calculate velocity within the specified portion, using definite-integrals.

Response to comment

V = delta(x)/delta(t), gives you the average velocity over x, when the time taken to cover x is t. Your problem in understanding is that you are confining x to a "small portion", and calculating average velocity for the time required to cover the "entire distance"---you should rather confine t as well, then it shall be well to proceed.

  • $\begingroup$ Well the object did accelerate, and has a constant force, thus a constant acceleration. I assume based on the main formula above that the velocity of that area should be less, since the distance now is much smaller. I mean even without the acceleration... $\endgroup$
    – Pupil
    Commented Jun 9, 2014 at 7:07
  • $\begingroup$ @Key: your question doesn't say anything about the acceleration. Can you update your question to give all the known details about the motion. $\endgroup$ Commented Jun 9, 2014 at 7:30
  • $\begingroup$ I have added more information about acceleration. $\endgroup$
    – Pupil
    Commented Jun 9, 2014 at 7:37
  • $\begingroup$ Abstract, I have confined both $X$ & $t$, so that would show the velocity is indeed less. Example: $A$ to $B$ distance is $10$ $m$, while area X distance is $4$ $m$, time of $A$ to $B$ is $3$ $seconds$, while time in $area$ $X$ is 1.5 seconds, by plugging those values in, $10/3$ or $4/1.5$ we see that the velocity is different, and yes we confined both $X$ and $t$. $\endgroup$
    – Pupil
    Commented Jun 17, 2014 at 18:12

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