Very basic question: When to use $s=vt$, $s=1/2vt$, $s=at$ and $s=a/t^2$?

Question

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?

Answer 1

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.

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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

Answer 2

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.

Answer 3

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 $[L]$.