# The dimensional formula of distance travelled in $n$th second

I read that the dimensional formula of distance traveled in $n^{th}$ second is same as that of velocity. Okay, the formula for the distance traveled in $n^{th}$ second is $s_t= u+\frac{a}{2}(2t-1)$ where $u$ is initial velocity, $a$ is uniform acceleration and $t$ is the time. If we proceed and expand the aforementioned formula we will get $s_t = u + at-\frac{a}{2}$. Now, the last term i.e. $-\frac{a}{2}$ is not a velocity and isn't the principle of homogeneity violated?

• My translation of Mr. Rennie's comment: An equation MUST be dimensionally consistent across the equal sign if it is to have any chance of being correct. Units MATTER in physics, and this is one area where I consistently have difficulty getting the concept across in my AP physics classes (I teach high school physics). If the equation is dimensionally inconsistent, it is JUST PLAIN INCORRECT! Aug 3, 2015 at 23:16

The distance travelled in a time $t$ is:

$$s = ut + \tfrac{1}{2}at^2$$

So the distance travelled between $t$ and $t - \Delta t$ is:

\begin{align} \Delta s &= s(t) - s(t - \Delta t) \\ &= ut + \tfrac{1}{2}at^2 - u(t - \Delta t) - \tfrac{1}{2}a(t - \Delta t)^2 \\ &= u\Delta t + \tfrac{1}{2}a(2t\Delta t - \Delta t^2) \end{align}

The equation you cite is obtained by setting $\Delta t = 1$, but remember that you're setting $\Delta t$ equal to one second not the dimensionless quantity $1$. So your equation should really be:

$$\Delta s = u\cdot (1 \space\text{second}) + \tfrac{1}{2}a(2t(1 \space\text{second}) - (1 \space\text{second})^2)$$

or multiplying this out:

$$\Delta s = u\cdot (1 \space\text{second}) + at(1 \space\text{second}) - \tfrac{1}{2}a(1 \space\text{second})^2$$

So it is dimensionally consistent.

It is dimensionally correct...

In the derivation, you take ut-u(t-1) which makes it appear as u, but is actually u*1 second...