Question

In the expression $v=a+\dfrac{b}{t+c}$ find the dimensional for a, b, c where v is the velocity, t is the time?

Answer 1

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 on 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 solution:
The given equation is
$\Rightarrow v=a+\dfrac{b}{t+c}$
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 travelled per time, we have the dimension of velocity as
$\Rightarrow \left[ v \right]=\left[ {{L}^{1}}{{T}^{-1}} \right]$
Now taking the right-hand side each quantity is being added to each other and the sum is equal to velocity. So, that the dimension of each term on the right-hand side will be equal to the dimension of velocity. Thus, we have
$\Rightarrow \left[ a \right]=\left[ v \right]=\left[ {{L}^{1}}{{T}^{-1}} \right]$
It is given that t denotes time and hence the dimension is $\left[ T \right]$ So
Similarly for the second term let us first consider the variable c. As it is being added with t of dimension $\left[ T \right]$, the dimension of c will also be the same. We have
$\Rightarrow \left[ c \right]=\left[ t \right]=\left[ T \right]$
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
$\Rightarrow \dfrac{\left[ b \right]}{\left[ T \right]}=\left[ {{L}^{1}}{{T}^{-1}} \right]$
$\Rightarrow \left[ b \right]=\left[ {{L}^{1}} \right]$
Therefore, for a dimensionally valid equation $v=a+\dfrac{b}{t+c}$ the dimension of a is $\left[ {{L}^{1}}{{T}^{-1}} \right]$ the dimension of b is $\left[ {{L}^{1}} \right]$ and the dimension of c is $\left[ T \right]$

Note: Since the velocity is defined as distance travelled per time, we have the dimension of velocity as $\left[ {{L}^{1}}{{T}^{-1}} \right]$. Students should not be confused about the dimension of c because it is added to the t that is dimension $\left[ T \right]$ so it also has the same dimension of t. Instead of this, if a student is supposed to equate the dimension of c with the dimension of v then the whole answer will be wrong.