If the velocity of a particle is `v = At + Bt^2`, where `A` and `B` are constant, then the distance travelled by it between `1 s` and `2 s` is :

Options

A. `3A+7B`
B. `(3)/(2)A+(7)/(3)B`
C. `(A)/(2)+(B)/(3)`
D. `(3)/(2)A+4B`

Explanation

To solve the problem, we need to find the distance traveled by a particle with a given velocity function \( v(t) = At + Bt^2 \) between \( t = 1 \) second and \( t = 2 \) seconds.

### Step-by-Step Solution:

1. **Understand the relationship between velocity and displacement**:
The velocity \( v \) is defined as the derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Therefore, we can write:
\[ \frac{dx}{dt} = At + Bt^2 \]

2. **Rearrange the equation**:
We can rearrange this to express \( dx \):
\[ dx = (At + Bt^2) dt \]

3. **Set up the integral for displacement**:
To find the total displacement \( \Delta x \) between \( t = 1 \) s and \( t = 2 \) s, we integrate \( dx \) from \( t = 1 \) to \( t = 2 \):
\[ \Delta x = \int_{1}^{2} (At + Bt^2) dt \]

4. **Break down the integral**:
We can split the integral into two parts:
\[ \Delta x = \int_{1}^{2} At \, dt + \int_{1}^{2} Bt^2 \, dt \]

5. **Calculate the first integral**:
The first integral is:
\[ \int At \, dt = \frac{A}{2} t^2 \Big|_{1}^{2} = \frac{A}{2} (2^2 - 1^2) = \frac{A}{2} (4 - 1) = \frac{3A}{2} \]

6. **Calculate the second integral**:
The second integral is:
\[ \int Bt^2 \, dt = \frac{B}{3} t^3 \Big|_{1}^{2} = \frac{B}{3} (2^3 - 1^3) = \frac{B}{3} (8 - 1) = \frac{7B}{3} \]

7. **Combine the results**:
Now, we can combine both results to find the total displacement:
\[ \Delta x = \frac{3A}{2} + \frac{7B}{3} \]

8. **Final expression**:
Thus, the distance traveled by the particle between \( t = 1 \) s and \( t = 2 \) s is:
\[ \Delta x = \frac{3A}{2} + \frac{7B}{3} \]