Find the dimension of (a/b) in the equation : v = a + bt , where v is velocity and t is time

Answer

To find the dimension of \( \frac{a}{b} \) in the equation \( v = a + bt \), where \( v \) is velocity and \( t \) is time, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of velocity (v)**:
- Velocity is defined as the displacement per unit time. Its dimension is given by:
\[ [v] = [L^1 T^{-1}] \]
where \( L \) represents length and \( T \) represents time.

2. **Identify the dimensions of time (t)**:
- Time has the dimension:
\[ [t] = [T^1] \]

3. **Analyze the equation \( v = a + bt \)**:
- Since \( a \) and \( bt \) are being added, they must have the same dimensions. Therefore, we can equate their dimensions:
\[ [a] = [bt] \]

4. **Find the dimension of \( b \)**:
- From the equation \( bt \), we can express \( b \) in terms of \( a \) and \( t \):
\[ [bt] = [b][t] \]
- Substituting the dimension of \( t \):
\[ [bt] = [b][T^1] \]
- Since \( [bt] = [v] = [L^1 T^{-1}] \), we can write:
\[ [b][T^1] = [L^1 T^{-1}] \]
- Rearranging gives us:
\[ [b] = \frac{[L^1 T^{-1}]}{[T^1]} = [L^1 T^{-2}] \]

5. **Find the dimension of \( a \)**:
- Since \( [a] = [bt] \), we can substitute the dimension of \( b \):
\[ [a] = [b][T^1] = [L^1 T^{-2}][T^1] = [L^1 T^{-1}] \]

6. **Calculate the dimension of \( \frac{a}{b} \)**:
- Now we can find the dimension of \( \frac{a}{b} \):
\[ \left[\frac{a}{b}\right] = \frac{[a]}{[b]} = \frac{[L^1 T^{-1}]}{[L^1 T^{-2}]} \]
- This simplifies to:
\[ \left[\frac{a}{b}\right] = \frac{[L^1 T^{-1}]}{[L^1 T^{-2}]} = [L^0 T^{1}] = [T^1] \]

### Final Result:
The dimension of \( \frac{a}{b} \) is:
\[ [T^1] \]