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

v=at+bt+c+v0v=at+\dfrac{b}{t+c} + v_0 is a dimensionally valid equation. 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

Use the fact the only dimensionally identical quantities can be operated upon together, which means that in an equation, every term of the RHS has the same dimensions as every term on the LHS. From the dimensions of velocity, solve for the dimensions of the required constants using the dimensions of time to simplify, and arrive at the appropriate solution.

Formula Used:
[v]=[L1T−1]\left[v\right] = \left[L^1T^{-1}\right]
[t]=[T1]\left[t\right] = \left[T^1\right]

Complete Step-by-Step Solution:
We are given an equation: v=at+bt+c+v0v=at+\dfrac{b}{t+c} + v_0

Since this is a dimensionally valid equation, it means that the dimensions of every term on LHS will be equal to the dimensions of the term on RHS.

We know that the dimensional formula of velocity [v]=[L1T−1]\left[v\right] = \left[L^1T^{-1}\right].

This means that the dimensional formula of initial velocity is also [v0]=[L1T−1]\left[v_0\right] = \left[L^1T^{-1}\right].

The dimensional formula of time is [t]=[T1]\left[t\right] = \left[T^1\right].

Using the above deductions, we now find the dimensional formula for a, b and c.
∙\bullet For a:

[at]=[v]⇒[a][t]=[v]⇒[a]=[v][t]\left[at\right] =\left[v\right] \Rightarrow \left[a\right] \left[t\right] = \left[v\right] \Rightarrow \left[a\right] = \dfrac{\left[v\right]}{ \left[t\right]}

⇒[a]=[L1T−1][T1]=[L1T−2]\Rightarrow \left[a\right] = \dfrac{\left[L^1T^{-1}\right]}{ \left[T^1\right]} = \left[L^1T^{-2}\right]

In the standard form, [a]=[L1M0T−2]\left[a\right]= \left[L^1M^0T^{-2}\right]

∙\bullet For c:

Since only dimensionally identical quantities can be added together, we can obtain the dimensions of c by considering the denominator of the term bt+c\dfrac{b}{t+c}.
⇒[c]=[t]=[T1]\Rightarrow \left[c\right] = \left[t\right] = \left[T^1\right]

In the standard form, this would be [c]=[L0M0T1]\left[c\right] = \left[L^0M^0T^1\right]

∙\bullet For b:
[bt+c]=[v]⇒[b][t+c]=[v]\left[\dfrac{b}{t+c}\right] = \left[v\right] \Rightarrow \dfrac{\left[b\right]}{ \left[t+c\right]} = \left[v\right]

[b][T1]=[LT−1]⇒[b]=[L1T(−1+1)]=[L1]\dfrac{\left[b\right]}{ \left[T^1\right]} = \left[LT^-1\right] \Rightarrow \left[b\right] = \left[L^1T^{(-1+1)}\right] = \left[L^1\right]

In the standard form, this would be [b]=[L1M0T0]\left[b\right] = \left[L^1M^0T^0\right]

Note:
The dimensional formula allows us to express the unit of a physical quantity in terms of the fundamental quantities mass (M), length (L), time (T) and temperature (K) in a standard form. We did not use temperature since they are most often concerned with thermodynamic equations.
However, while deducing the dimensional formula of the constants from the base units be careful in assigning the right power to the right fundamental quantity each time as any discrepancy in the dimension will result in an anomalous nature of the quantities involved.