The velocity of a particle change with time according to the relation `v=xt+yt^(2)+z`. Give the dimensions of x,y and z if v is in `ms^(-1)` and t is in s.

Explanation

To solve the problem, we need to determine the dimensions of the constants \( x \), \( y \), and \( z \) in the equation \( v = xt + yt^2 + z \), given that \( v \) is in \( \text{m/s} \) and \( t \) is in seconds.

### Step-by-Step Solution:

1. **Identify the dimensions of velocity \( v \)**:
- The velocity \( v \) is given in meters per second (m/s).
- Therefore, the dimension of \( v \) can be expressed as:
\[ [v] = L T^{-1} \]
where \( L \) represents length and \( T \) represents time.

2. **Analyze the equation \( v = xt + yt^2 + z \)**:
- According to the principle of homogeneity of dimensions, all terms on the right-hand side must have the same dimensions as the left-hand side.
- Thus, each term \( xt \), \( yt^2 \), and \( z \) must also have the dimensions of \( L T^{-1} \).

3. **Determine the dimensions of \( x \)**:
- For the term \( xt \):
\[ [xt] = [x][t] \]
- Since \( [t] = T \), we have:
\[ [xt] = [x] T \]
- Setting this equal to the dimension of \( v \):
\[ [x] T = L T^{-1} \]
- Rearranging gives:
\[ [x] = L T^{-1} \cdot T^{-1} = L T^{-2} \]

4. **Determine the dimensions of \( y \)**:
- For the term \( yt^2 \):
\[ [yt^2] = [y][t^2] \]
- Since \( [t^2] = T^2 \), we have:
\[ [yt^2] = [y] T^2 \]
- Setting this equal to the dimension of \( v \):
\[ [y] T^2 = L T^{-1} \]
- Rearranging gives:
\[ [y] = L T^{-1} \cdot T^{-2} = L T^{-3} \]

5. **Determine the dimensions of \( z \)**:
- For the term \( z \):
\[ [z] = L T^{-1} \]
- This is already equal to the dimension of \( v \), so we do not need any further calculations.

### Final Dimensions:
- The dimensions of the constants are:
- \( [x] = L T^{-2} \)
- \( [y] = L T^{-3} \)
- \( [z] = L T^{-1} \)

### Summary:
- \( [x] = L T^{-2} \)
- \( [y] = L T^{-3} \)
- \( [z] = L T^{-1} \)