Question

Check the correctness of the following equation by the method of dimensions:
$S = ut + \dfrac{1}{2}a{t^2}$
where $S$ is the distance covered by a body in time $t$, having initial velocity $u$ and acceleration $a$.
A) At LHS=RHS, the formula is dimensionally correct
B) LHS and RHS doesn’t have the same dimension, hence incorrect
C) RHS is dimensionally correct
D) None of these

Answer 1

In this solution, we will check the formula of distance provided to us by checking whether the terms have appropriate dimensional formulae or not. For two terms to be added, they need to have the same dimensional formula.

Complete step by step answer
We’ve been asked to check the equation $S = ut + \dfrac{1}{2}a{t^2}$ for its dimensions. For the left side and the right side to be dimensionally correct, both sides need to have the same dimensional formula.
The term on the left side will have the dimensions of distance i.e.
$[S] = {L^1}$
The terms on the right side are two separate terms. In the first term, we have the product of velocity and time so,
$[ut] = {L^1}{T^{ - 1}} \times {T^1}$
$\therefore [ut] = {L^1}$
The second term on the right side will have the product of acceleration and square of time. Since the dimensions of acceleration are $[a] = {L^1}{T^{ - 2}}$, the dimensions of the second term on the right side will be:
$\left[ {\dfrac{1}{2}a{t^2}} \right] = \left( {{L^1}{T^{ - 2}}} \right)\left( {{T^2}} \right)$
$\Rightarrow \left[ {\dfrac{1}{2}a{t^2}} \right] = {L^1}$
So, both terms have the same dimensions of length. So, we can add both the terms on the right side and the sum will still have the dimensions of length. Since both the sides have the dimensions of length, the equation is dimensionally correct.

Hence the correct option is option (A).

Note
The equation given to us is the second equation of kinematics which relates the distance travelled by an object for an initial given velocity experiencing constant acceleration. While the equation given to us is physically correct, an equation can be dimensionally correct without being physically correct. So we should only focus on the dimensional formula of the equation in this case since that’s what we’ve been asked about.