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 \( b \) in the equation \( v = at + bt^2 \), we can follow these steps:

### Step 1: Understand the equation
The equation given is:
\[ v = at + bt^2 \]
Here, \( v \) is the velocity of the particle, \( a \) is a constant, \( b \) is another constant, and \( t \) is time.

### Step 2: Identify the dimensions of velocity
The dimension of velocity \( v \) is given by:
\[ [v] = L T^{-1} \]
where \( L \) represents length and \( T \) represents time.

### Step 3: Analyze the terms in the equation
Since \( v \) is expressed as \( at + bt^2 \), both terms \( at \) and \( bt^2 \) must have the same dimensions as \( v \).

### Step 4: Determine the dimensions of \( at \)
The term \( at \) has dimensions:
\[ [a] \cdot [t] = [a] \cdot T \]
Thus, the dimension of \( a \) can be expressed as:
\[ [a] = \frac{[v]}{[t]} = \frac{L T^{-1}}{T} = L T^{-2} \]

### Step 5: Determine the dimensions of \( bt^2 \)
Now, we analyze the term \( bt^2 \):
\[ [b] \cdot [t^2] = [b] \cdot T^2 \]
Since \( bt^2 \) must also have the same dimension as \( v \):
\[ [b] \cdot T^2 = L T^{-1} \]

### Step 6: Solve for the dimension of \( b \)
Rearranging the equation gives:
\[ [b] = \frac{L T^{-1}}{T^2} = L T^{-3} \]
Thus, the dimension of \( b \) is:
\[ [b] = L T^{-3} \]

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

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