Question

The velocity of a body is \[{\text{v = A + Bt}}\] , then what are the dimensions of \[{\text{A}}\] and \[{\text{B}}\] ?

Answer 1

Hint: Dimensions of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity. The principle of Homogeneity of Dimension states that the dimensions of all the terms in a physical expression should be the same. Thus, in the expression \[{\text{v = A + Bt}}\] dimension of \[{\text{v}}\] , \[{\text{A}}\] and \[{\text{Bt}}\] are all same.

Complete step by step answer:
We are given, the velocity of a body can be expressed as \[{\text{v = A + Bt}}\].
Velocity is defined as \[{\text{v}} = \dfrac{\text{displacement}}{\text{time}}\]
Thus, the dimension of velocity is,
\[{\text{v}} = \dfrac{{\left[ L \right]}}{{\left[ T \right]}} = \left[ {{M^0}L{T^{ - 1}}} \right]\]
The principle of Homogeneity of Dimension states that the dimension of all the terms in a physical relation should be the equal. Using The principle of Homogeneity of Dimension we can say, dimension of A is equal to the dimension of velocity. So, we get,
\[\left[ A \right] = \left[ {\text{v}} \right]\]
\[ \Rightarrow \left[ A \right] = \left[ {{M^0}L{T^{ - 1}}} \right]\]
Dimension of \[{\text{Bt}}\] is equal to the dimension of velocity
\[ \Rightarrow \left[ {Bt} \right] = \left[ {{M^0}L{T^{ - 1}}} \right]\]
\[ \Rightarrow \left[ B \right] = \dfrac{{\left[ {{M^0}L{T^{ - 1}}} \right]}}{{\left[ T \right]}}\]
\[ \therefore \left[ B \right] = \left[ {{M^0}L{T^{ - 2}}} \right]\]

Thus, the dimension of \[{\text{A}}\] is \[\left[ {{M^0}L{T^{ - 1}}} \right]\] and dimension of \[{\text{B}}\] is \[\left[ {{M^0}L{T^{ - 2}}} \right]\].

Additional information: Dimensional analysis is an important aspect of measurement and has the following applications:
To check the correctness of an equation or the other physical relation supports the principle of homogeneity. There should be dimensions on two sides of the equation. The dimensional relation is going to be correct if the L.H.S and R.H.S of an equation have identical dimensions. If the dimensions on two sides are incorrect, then the relations will also be incorrect accordingly.

Dimensional analysis is employed to convert the worth of a physical quantity from one system of units to a different system of units. Dimensional analysis is employed to represent the character of physical quantity. The expressions of dimensions are often manipulated as algebraic quantities. Dimensional analysis is used to derive formulas.

Note: Dimensional analysis doesn't give information about the scalar or vector identity of a physical quantity. We cannot derive trigonometric, exponential, and logarithmic functions using dimensional analysis. It doesn't provide information about the dimensional constant.