Question

If $x = a + bt + c{t^2}$ , where $x$ is in meters and $t$ is in second. What is the unit of $b$ and $c$?

Answer 1

Examine the question, this is from the unit and dimension topic. This is particularly the law of homogeneity. If x is in meter then according to law of homogeneity ${\text{a, bt}},{\text{ c}}{{\text{t}}^2}$ have units of meter. Law of homogeneity states that the dimensions of each the terms of a dimensional equation on both sides are the same.

Complete step by step answer:
Given equation $x = a + bt + c{t^2}$ ,
Here, units of x are meters.
Therefore by the law of homogeneity,
Units of x are the same as units of a.
$\Rightarrow$ units of a $=$ Meter (m)
units of $x$ $=$ units of $a$ = units of $bt$ = units of $c{t^2}$
from above:
$x = {\text{ a = [}}{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^0}{\text{]}}$
$x{\text{ = bt }} \\\ \Rightarrow {\text{ b = }}\dfrac{x}{t}{\text{ = }}\dfrac{{{\text{[}}{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^0}]}}{{{\text{[}}{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^1}{\text{]}}}}{\text{ = [}}{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 1}}{\text{]}}$
Therefore units of $b$ is $ms^{-1}$. Similarly,
$x{\text{ = c}}{{\text{t}}^2}{\text{ }} \\\ \Rightarrow {\text{ c = }}\dfrac{x}{{{t^2}}}{\text{ = }}\dfrac{{{\text{[}}{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^0}]}}{{{\text{[}}{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^2}{\text{]}}}}{\text{ = [}}{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 2}}{\text{]}}$

Therefore the unit of c is $ms^{-2}$.

Note: It is necessary to find dimensions of any one term to apply the law of homogeneity. Find the dimensional formula for each term in the equation to find the dimensions of the required variable. To find the dimensional formula analyse the question carefully. Try to solve these questions in SI units to avoid confusions.