Question

if the the displacement of
particles in the the t th second
is S = U+a/2( 2t-1)find the dimension of
S from your
observation​

Answer 1

Since all are related to each to each by = or +, their dimensions(dimension formula) must be 0. If we just look at this at a glance, we would write:

[S] = [u] = [a/2 (2t - 1)]

[L] = [LT-¹] = [LT-¹]

Ignoring the facts, you can say dimension formula of S is [LT-¹] and dimensions are 0, 1, -1.

*which is totally false.

This would make you believe that [S] is [LT-¹]. It means, we say distance = speed.

Note that: displacement in nth is:

= S(n) - S(n-1)

= un + ½ at² - [un - 1) + ½ a(t - 1)²]

Here, since 1 is subtracted from t, 1(unit) must have same dimensions i.e. [T].

Hence, the eqⁿ is actually,

S = u*(1s) + (a/2)(2t - 1)*(1s)

For more simplicity:

You have nth sec formula for 1second. What if we need the displacement in nth + (n + 1)th sec? Or something like displacement in 2sec, 3sec, 4sec, etc. Here is a more general formula:

Let, you need to find displacement for last t second, currently at time = n.

= S(n) - S(n - t)

= un + ½ an² - [u(n - t) + ½ a(n - t)²]

= un + ½ an² - [un - ut + ½ (an² + at² -2atn] = ut + ½ (2atn - at²)

= ut + (a/2) (2n - t)t

It even satisfies for 1 sec, t = 1

S(nth) = u(1) + (a/2) (2n - 1)1 = u + (a/2)(2n- 1)

S(for last t sec) = ut + (a/2) (2n - t)t

is better form to be used.

Now, if you check the dimensions

[S] = [ut] = [(a/2)(2n - t)t]

[L] = [LT-¹ T] = [(LT-²)(T)(T)]

[L] = [L] = [L] , which is true.

Hence,

dimension formula of S is [L].