`( alpha)/( t^(2)) = Fv + ( beta)/(x^(2))`
Find the dimension formula for `[ alpha] and [ beta]` ( here t = time , F = force , v = velocity , x = distance).

Explanation

To find the dimension formulas for \([ \alpha ]\) and \([ \beta ]\) from the equation

\[ \frac{\alpha}{t^2} = Fv + \frac{\beta}{x^2} \]

we will follow these steps:

### Step 1: Identify the dimensions of the variables involved

1. **Force (F)**: The dimension of force is given by:
\[ [F] = [M][L][T^{-2}] \]
where \([M]\) is mass, \([L]\) is length, and \([T]\) is time.

2. **Velocity (v)**: The dimension of velocity is given by:
\[ [v] = [L][T^{-1}] \]

3. **Time (t)**: The dimension of time is:
\[ [t] = [T] \]

4. **Distance (x)**: The dimension of distance is:
\[ [x] = [L] \]

### Step 2: Calculate the dimension of \(Fv\)

Now, we calculate the dimension of the product \(Fv\):
\[ [Fv] = [F] \cdot [v] = ([M][L][T^{-2}]) \cdot ([L][T^{-1}]) = [M][L^2][T^{-3}] \]

### Step 3: Set up the equation for \(\alpha\)

From the equation, we have:
\[ \frac{\alpha}{t^2} = Fv \]
This implies:
\[ [\alpha] = [Fv] \cdot [t^2] \]
Substituting the dimensions we found:
\[ [\alpha] = [M][L^2][T^{-3}] \cdot [T^2] = [M][L^2][T^{-1}] \]

### Step 4: Calculate the dimension of \(\beta\)

Next, we analyze the term \(\frac{\beta}{x^2}\):
\[ \frac{\beta}{x^2} = Fv \]
This implies:
\[ [\beta] = [Fv] \cdot [x^2] \]
Substituting the dimensions we found:
\[ [\beta] = [M][L^2][T^{-3}] \cdot [L^2] = [M][L^4][T^{-3}] \]

### Final Results

Thus, the dimension formulas for \(\alpha\) and \(\beta\) are:
\[ [\alpha] = [M][L^2][T^{-1}] \]
\[ [\beta] = [M][L^4][T^{-3}] \]