The velocity (V) of a particle (in cm/s) is given in terms of time (t) in sec by the equation `V=at+(b)/(c+t)`. The dimensions of a, b and c are

Options

A. `{:(a,b,c),(L^(2),T,L^(1)T^(-2)):}`
B. `{:(a,b,c),(L^(1)T^(2),L^(1)T,L^(1)):}`
C. `{:(a,b,c),(L^(1)T^(-2),L^(1),T^(1)):}`
D. `{:(a,b,c),(L^(1),L^(1)T,T^(2)):}`

Explanation

To solve the problem, we need to determine the dimensions of the variables \(a\), \(b\), and \(c\) from the given equation for velocity \(V\):

\[ V = at + \frac{b}{c + t} \]

### Step 1: Identify the dimensions of \(V\)

The velocity \(V\) is given in centimeters per second (cm/s). Therefore, the dimensions of \(V\) can be expressed as:

\[ [V] = L T^{-1} \]

where \(L\) represents length and \(T\) represents time.

### Step 2: Determine the dimensions of \(a\)

From the equation, we see that \(V = at\). Rearranging gives us:

\[ a = \frac{V}{t} \]

Substituting the dimensions of \(V\) and \(t\):

\[ [a] = \frac{L T^{-1}}{T} = L T^{-2} \]

### Step 3: Analyze the term \(\frac{b}{c + t}\)

Next, we need to analyze the term \(\frac{b}{c + t}\). For this term to be valid, the dimensions of \(b\) must match the dimensions of \(V\) when divided by \(c + t\).

Since \(t\) has dimensions of \(T\), the dimensions of \(c\) must also be \(T\) for the addition \(c + t\) to be dimensionally consistent. Thus, we have:

\[ [c] = T \]

### Step 4: Set up the equation for \(b\)

Now we can express the dimensions of \(b\):

\[ \frac{b}{c + t} = V \]

This implies:

\[ [b] = V \cdot (c + t) \]

Substituting the dimensions we have:

\[ [b] = (L T^{-1}) \cdot T = L T^{0} = L \]

### Summary of dimensions

Now, we can summarize the dimensions of \(a\), \(b\), and \(c\):

- \( [a] = L T^{-2} \)
- \( [b] = L \)
- \( [c] = T \)

### Final Answer

Thus, the dimensions of \(a\), \(b\), and \(c\) are:

- \(a\): \(L T^{-2}\)
- \(b\): \(L\)
- \(c\): \(T\)