Solveeit Logo
Login

Question

Question: A dimensionally valid equation is \(v = at + \dfrac{b}{{t + c}} + {v_0}\). Obtain the dimensional fo...

A dimensionally valid equation is v=at+bt+c+v0v = at + \dfrac{b}{{t + c}} + {v_0}. Obtain the dimensional formula for a, b, and c where, v is the velocity, t is the time and v0v_0 is the initial velocity.

Explanation

Solution

If two quantities are to be added with each other or subtracted from one another, the two quantities have to be dimensionally identical. That is, each term in the right hand side of the equation should be matching with that of the term on the left hand side in terms of dimensions.

Complete step by step answer:
The given equation is v=at+bt+c+v0v = at + \dfrac{b}{{t + c}} + {v_0}. It is said that the equation is valid in terms of dimensions, that means, the dimensions of the terms on the left hand side will be equal to the dimensions of the terms on the right hand side. Also, it is given that the term on the left hand side, that is v denotes velocity. Since the velocity is defined as distance traveled per time, we have the dimension of velocity as,
[v]=[L1T−1][v] = [{L^1}{T^{ - 1}}]

Now taking the right hand side, each quantity is being added to each other and the sum is equal to velocity. So it can be inferred that the dimension of each term on the right hand side will be equal to the dimension of velocity. Thus, we have,
[at]=[v]=[L1T−1][at] = [v] = [{L^1}{T^{ - 1}}]
It is given that t denotes time and hence the dimension is, [T]. So,
[a][t]=[L1T−1]         [a]=[L1T−1][T] =[L1T−2] [a][t] = [{L^1}{T^{ - 1}}]\\\ \;\;\;\;[a] = \dfrac{{[{L^1}{T^{ - 1}}]}}{{[T]}}\\\ = [{L^1}{T^{ - 2}}]
Similarly for the second term let us first consider the variable c. As it is being added with t of dimension [T], the dimension of c will also be the same. We have,
[c]=[t]=[T1][c] = [t] = [{T^1}]
Now considering again the second term, we have the dimension of the denominator, and the whole term as such will have the dimension equal to that of velocity. Therefore we have,
[b][T]=[L1T−1] [b]=[L1] \dfrac{{[b]}}{{[T]}} = [{L^1}{T^{ - 1}}]\\\ [b] = [{L^1}]
Therefore, for a dimensionally valid equationv=at+bt+c+v0v = at + \dfrac{b}{{t + c}} + {v_0}, the dimension of a is [L1T−2][L^1 T^{-2}], the dimension of b is [L1][L^1] and the dimension of c is [T1].[T^1].

Note:
Since the velocity is defined as distance traveled per time, we have the dimension of velocity as, [L1T−2].[L^1T^{-2}].
The last term is given as initial velocity and it has the same dimension as that of velocity
The dimensions of the terms on the left hand side will be equal to the dimensions of the terms on the right hand side.