Determining Dimensions of Constants in Velocity Equation
We are given the velocity $v$ of a particle at time $t$ as: $v = at + \frac{b}{t+c}$ We need to find the dimensions of the constants $a$, $b$, and $c$. We will use the principle of dimensional homogeneity, which states that dimensions must be consistent throughout an equation.
Dimensional Analysis Steps
- Dimension of Velocity ($v$): Velocity is defined as displacement over time. Therefore, its dimensions are length per time. $[v] = LT^{-1}$
- Dimension of Constant $a$: The term $at$ must have the same dimensions as the velocity $v$, since it's added to other terms contributing to $v$. $[at] = [v]$ $[a][t] = LT^{-1}$ Since the dimension of time $t$ is $[t] = T$, we substitute this: $[a]T = LT^{-1}$ Solving for $[a]$: $[a] = \frac{LT^{-1}}{T} = LT^{-2}$
- Dimension of Constant $c$: In the term $t+c$, both $t$ and $c$ must have the same dimensions because quantities can only be added or subtracted if they have identical dimensions. $[c] = [t]$ $[c] = T$
- Dimension of Constant $b$: The term $\frac{b}{t+c}$ must also have the same dimensions as velocity $v$. $\left[\frac{b}{t+c}\right] = [v]$ $\frac{[b]}{[t+c]} = LT^{-1}$ From step 3, we know $[t+c] = [t] = T$. Substituting this: $\frac{[b]}{T} = LT^{-1}$ Solving for $[b]$: $[b] = LT^{-1} \times T = L$
Resulting Dimensions
The dimensions are found to be:
- Dimension of $a$: $LT^{-2}$
- Dimension of $b$: $L$
- Dimension of $c$: $T$
This matches option 3.
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