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

Explanation

To find the distance travelled by a particle between 1 second and 2 seconds when its velocity is given by \( v = At + Bt^2 \), we can follow these steps:

### Step 1: Understand the relationship between velocity and distance
The velocity \( v \) is the derivative of the distance \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Given that \( v = At + Bt^2 \), we can write:
\[ \frac{dx}{dt} = At + Bt^2 \]

### Step 2: Integrate the velocity to find the distance
To find the distance \( x \) travelled between \( t = 1 \) second and \( t = 2 \) seconds, we need to integrate the velocity function with respect to time:
\[ x = \int (At + Bt^2) \, dt \]

### Step 3: Perform the integration
The integral of \( At + Bt^2 \) is:
\[ x = \int (At + Bt^2) \, dt = \frac{A}{2}t^2 + \frac{B}{3}t^3 + C \]
where \( C \) is the constant of integration. However, since we are interested in the distance travelled between two specific times, we can ignore \( C \) for now.

### Step 4: Evaluate the definite integral from 1 to 2 seconds
We need to evaluate the distance from \( t = 1 \) to \( t = 2 \):
\[ x(2) - x(1) = \left( \frac{A}{2}(2^2) + \frac{B}{3}(2^3) \right) - \left( \frac{A}{2}(1^2) + \frac{B}{3}(1^3) \right) \]

Calculating \( x(2) \):
\[ x(2) = \frac{A}{2}(4) + \frac{B}{3}(8) = 2A + \frac{8B}{3} \]

Calculating \( x(1) \):
\[ x(1) = \frac{A}{2}(1) + \frac{B}{3}(1) = \frac{A}{2} + \frac{B}{3} \]

### Step 5: Subtract the two results
Now, we compute the distance travelled:
\[ \text{Distance} = x(2) - x(1) = \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right) \]

### Step 6: Simplify the expression
Combining the terms:
\[ = 2A - \frac{A}{2} + \frac{8B}{3} - \frac{B}{3} \]
\[ = \left( 2A - \frac{A}{2} \right) + \left( \frac{8B}{3} - \frac{B}{3} \right) \]
\[ = \left( \frac{4A}{2} - \frac{A}{2} \right) + \left( \frac{8B - B}{3} \right) \]
\[ = \frac{3A}{2} + \frac{7B}{3} \]

### Final Result
Thus, the distance travelled by the particle between 1 second and 2 seconds is:
\[ \text{Distance} = \frac{3A}{2} + \frac{7B}{3} \]