Position of two particles A and B as a function of time are given by `X_A = – 3t^2 + 8t + c` and `Y_B = 10 – 8t^3`. The velocity of B with respect to A at `t = 1` is `sqrt(v)` . Find v.

Explanation

To find the value of \( v \) given the positions of two particles A and B, we will follow these steps:

### Step 1: Determine the velocity of particle A
The position of particle A is given by:
\[ X_A = -3t^2 + 8t + c \]
To find the velocity of particle A, we differentiate \( X_A \) with respect to time \( t \):
\[ V_A = \frac{dX_A}{dt} = \frac{d}{dt}(-3t^2 + 8t + c) = -6t + 8 \]
Now, we will evaluate the velocity at \( t = 1 \):
\[ V_A(1) = -6(1) + 8 = -6 + 8 = 2 \, \text{m/s} \]

### Step 2: Determine the velocity of particle B
The position of particle B is given by:
\[ Y_B = 10 - 8t^3 \]
To find the velocity of particle B, we differentiate \( Y_B \) with respect to time \( t \):
\[ V_B = \frac{dY_B}{dt} = \frac{d}{dt}(10 - 8t^3) = -24t^2 \]
Now, we will evaluate the velocity at \( t = 1 \):
\[ V_B(1) = -24(1^2) = -24 \, \text{m/s} \]

### Step 3: Calculate the velocity of B with respect to A
The velocity of B with respect to A is given by:
\[ V_{BA} = V_B - V_A \]
Substituting the values we found:
\[ V_{BA} = -24 - 2 = -26 \, \text{m/s} \]
The magnitude of the velocity of B with respect to A is:
\[ |V_{BA}| = 26 \, \text{m/s} \]

### Step 4: Relate the velocity to the given expression
According to the problem, the velocity of B with respect to A is given as \( \sqrt{v} \):
\[ |V_{BA}| = \sqrt{v} \]
Thus, we have:
\[ 26 = \sqrt{v} \]

### Step 5: Solve for \( v \)
To find \( v \), we square both sides:
\[ v = 26^2 = 676 \]

### Final Answer
The value of \( v \) is:
\[ \boxed{676} \]