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Very basic question: When to use $s=vt$, $s=\frac{1}{2}vt$, $s=at$ and $s=\frac{a}{t^2}$? What was the difference between those?

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I've added LaTeX markup so that the MathJax engine will render you math nicely. You should probably proofread my work as I had to guess you intent. –  dmckee Oct 10 '11 at 13:57
    
Only the first one makes any sense dimensionally. When given an equation first check the dimensions (the units) and only if it works out ok proceed with trying to analyse how to use it and what to get out of it. –  ja72 Oct 11 '11 at 4:30
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Hi @dmckee: the equations should read $s=ut+\frac{at^2}{2}$, $s=\frac{(v+u)t}{2}$, $v=u+at$, and $v^2=u^2+2as$. Or at least they are the ones I was forced to remember some 10 years ago ;-) –  qftme Oct 11 '11 at 15:18
    
@Well, yes. But I rendered what the OP wrote with the best fidelity I could manage, because the oddness of the question is important to providing a useful answer. –  dmckee Oct 11 '11 at 15:21
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3 Answers

I will give you them one by one, here

A body is moving under a constant acceleration $a$ with $u$ as initial velocity, $v$ as final velocity in time $t$.

  1. $$s = vt$$ This is valid only for a system under constant velocity. i.e. when $a = 0$ .
  2. $$s = \frac{1}{2}vt$$ This is wrong. Since the correct one is item 1.
  3. $$s = at$$ This is too wrong since dimensionality is violated. The correct equation is $$s= ut + 1/2 \times at^2$$. where $u$ is initial velocity
  4. $$s = \frac{a}{t^2}$$ Wrong because of dimensionality.

I will give you three equation which might help you in solving kinematic problems.

  1. $$v=u+at$$
  2. $$s=ut+\frac{1}{2}at^2$$
  3. $$v^2=u^2+2as$$

all of these are quite simple to derive using basic principles.

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You forgot to say that $s$ is the distance. It is not the notation used anywhere in the world. –  Frédéric Grosshans Jun 15 '12 at 14:42
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@FrédéricGrosshans: I don't know about rest of the world, but in India, that's the practice... at least in engineering/high school... $s$ stands for displacement... –  Vineet Menon Jun 16 '12 at 15:02
    
+1 @VineetMenon: Hello Vineet, Yes. I agree with that practice... –  Waffle's Crazy Peanut Oct 28 '12 at 5:00
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Pay attention to the dimensions. The fundamental physical quantities used in mechanics are length, denoted by $[L]$, mass $[M]$, and time $[T]$. For equations to make sense the dimensions must be in agreement.

Imagine finding average speed, average speed = distance/time, but instead of speed having a length/time dimension you get mass. So it wouldn't make sense; it is not logical.

$$s=vt$$

$s$ is distance so it is a length quantity with dimension $[L]$, and velocity is distance/time thus it has dimensions

$[L]/[T]$, and so on.

Going back to $s = vt$,

$$[L] = [L]/[T] * [T]$$

Correct in dimension.

$$s=1/2vt$$

$$[L] = [L]/[T] * [T]$$

Correct in dimension. However, the magnitude of the right side does not follow the definition of velocity.

$$s=at$$

$$[L] = [L]/[T]^2 * [T]$$

Wrong. Notice that the right side of the equation is equal to $[L]/[T]$ and not $[L]$.

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The various basic kinematic equations were found under a certain set of assumptions.

For instance

$$ \text{distance} = \text{velocity} * \text{time} $$

was found under the assumption of constant velocity (i.e. zero acceleration), while

$$ s = \frac{1}{2} a t^2 $$

was found under the assumption of constant acceleration (and starting from rest, $v=0$, at $s=0$,$t=0$).

You use them only when the conditions in your problem match the ones under which they were found.


On a side note, the latter two that you list are not generally among those we derive for first year students and are not valid in any simple---but---general cases that I am aware of

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In fact, the latter two eqs. are wrong for dimensional reasons alone if $a$ is supposed to be acceleration ... –  Qmechanic Oct 10 '11 at 14:12
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If you take $v$ to mean the final velocity, and set $v_{i}=0$ under constant acceleration then I guess you could find the result $s=\frac{1}{2}vt={\bar v}t$. It would be a silly thing to memorize, of course. –  Jerry Schirmer Oct 10 '11 at 16:18
    
@Qmechanic id $s$ were for *s*peed, I could understand the 3rd equation... but I've no idea if this kind of notation is used in English. –  Frédéric Grosshans Jun 15 '12 at 14:40
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