During a short interval of time the speed v in m/s of an automobile is given by `v=at^(2)+bt^(3)`, where the time t is in seconds. The units of a and b are, respectively.

Options

A. `ms^(2),ms^(4)`
B. `s^(3)//m,s^(4)//m`
C. `m//s^(2),m//s^(3)`
D. `m//s^(3),m//s^(4)`

Answer

To determine the units of \( a \) and \( b \) in the equation \( v = at^2 + bt^3 \), where \( v \) is the speed in meters per second (m/s) and \( t \) is the time in seconds (s), we can follow these steps:

### Step 1: Identify the dimensions of speed \( v \)
The speed \( v \) is given in meters per second (m/s). The dimensional formula for speed is:
\[ [v] = L T^{-1} \]
where \( L \) represents length (meters) and \( T \) represents time (seconds).

### Step 2: Analyze the term \( at^2 \)
In the term \( at^2 \):
- The time \( t \) has dimensions of \( T \).
- Therefore, \( t^2 \) has dimensions of \( T^2 \).

Thus, the dimensions of \( at^2 \) can be expressed as:
\[ [at^2] = [a][t^2] = [a] T^2 \]
Since \( [v] = L T^{-1} \), we can equate:
\[ [a] T^2 = L T^{-1} \]

### Step 3: Solve for the dimensions of \( a \)
Rearranging the equation gives:
\[ [a] = \frac{L T^{-1}}{T^2} = L T^{-3} \]
Substituting the dimensions:
\[ [a] = L T^{-3} = \text{meters} \cdot \text{seconds}^{-3} \]

### Step 4: Analyze the term \( bt^3 \)
In the term \( bt^3 \):
- The time \( t \) has dimensions of \( T \).
- Therefore, \( t^3 \) has dimensions of \( T^3 \).

Thus, the dimensions of \( bt^3 \) can be expressed as:
\[ [bt^3] = [b][t^3] = [b] T^3 \]
Again, since \( [v] = L T^{-1} \), we can equate:
\[ [b] T^3 = L T^{-1} \]

### Step 5: Solve for the dimensions of \( b \)
Rearranging the equation gives:
\[ [b] = \frac{L T^{-1}}{T^3} = L T^{-4} \]
Substituting the dimensions:
\[ [b] = L T^{-4} = \text{meters} \cdot \text{seconds}^{-4} \]

### Conclusion
The units of \( a \) and \( b \) are:
- \( a \): \( \text{m} \cdot \text{s}^{-3} \)
- \( b \): \( \text{m} \cdot \text{s}^{-4} \)