If `x = at + bt^(2)`, where x is in meter and t in hours (hr) then the unit of b is

Options

A. m
B. m/hr
C. m/hr`""^(2)`
D. `m^(2)//hr`

Answer

To find the unit of \( b \) in the equation \( x = at + bt^2 \), where \( x \) is in meters and \( t \) is in hours, we will follow these steps:

### Step 1: Identify the units of each term
The left-hand side (LHS) of the equation is \( x \), which has the unit of meters (m). The right-hand side (RHS) consists of two terms: \( at \) and \( bt^2 \).

### Step 2: Analyze the term \( at \)
Let’s denote the unit of \( a \) as \( [a] \). Since \( t \) is in hours (hr), the unit of the term \( at \) will be:
\[ [a] \cdot \text{(hr)} \]
For the equation to be dimensionally consistent, the unit of \( at \) must also be meters (m):
\[ [a] \cdot \text{(hr)} = \text{m} \]
From this, we can express the unit of \( a \):
\[ [a] = \frac{\text{m}}{\text{hr}} \]

### Step 3: Analyze the term \( bt^2 \)
Now, let’s analyze the term \( bt^2 \). The unit of \( t^2 \) (since \( t \) is in hours) will be:
\[ \text{(hr)}^2 = \text{hr}^2 \]
Thus, the unit of the term \( bt^2 \) will be:
\[ [b] \cdot \text{(hr)}^2 \]
For the equation to be dimensionally consistent, the unit of \( bt^2 \) must also be meters (m):
\[ [b] \cdot \text{(hr)}^2 = \text{m} \]
From this, we can express the unit of \( b \):
\[ [b] = \frac{\text{m}}{\text{(hr)}^2} \]

### Conclusion
Thus, the unit of \( b \) is:
\[ \text{m} \cdot \text{hr}^{-2} \]

### Final Answer
The unit of \( b \) is \( \frac{\text{m}}{\text{hr}^2} \).

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