The velocity of a paritcle (v) at an instant t is given by `v=at+bt^(2)`. The dimesion of b is

Options

A. [L]
B. `[LT^(-1)]`
C. `[LT^(-2)]`
D. `[LT^(-3)]`

Explanation

To find the dimension of the constant \( b \) in the equation \( v = at + bt^2 \), we will follow these steps:

### Step 1: Identify the dimensions of velocity
The velocity \( v \) has dimensions of length per time. Therefore, we can express the dimensions of \( v \) as:
\[ [v] = L^1 T^{-1} \]

### Step 2: Analyze the equation
The equation given is:
\[ v = at + bt^2 \]
Here, \( a \) and \( b \) are constants, and \( t \) is time.

### Step 3: Determine the dimensions of \( at \)
Since \( v \) is the sum of \( at \) and \( bt^2 \), the dimensions of \( at \) must also equal the dimensions of \( v \).

The dimension of \( t \) (time) is:
\[ [t] = T^1 \]
Thus, the dimensions of \( a \) can be expressed as:
\[ [a] = \frac{[v]}{[t]} = \frac{L^1 T^{-1}}{T^1} = L^1 T^{-2} \]

### Step 4: Determine the dimensions of \( bt^2 \)
Now we will analyze \( bt^2 \). The dimension of \( t^2 \) is:
\[ [t^2] = T^2 \]
Therefore, the dimensions of \( b \) can be expressed as:
\[ [b] = \frac{[v]}{[t^2]} = \frac{L^1 T^{-1}}{T^2} = L^1 T^{-3} \]

### Step 5: Conclusion
Thus, the dimension of \( b \) is:
\[ [b] = L^1 T^{-3} \]

### Final Answer
The dimension of \( b \) is \( L^1 T^{-3} \).

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