The motion of a particle is described by x = `x_o(1 - e^(-kt))`,, t`ge`, `x_o`gt0, k gt 0. With what velocity does the particle start?

Options

A. `x_o`/k
B. `x_o`k
C. `k/x_o`
D. `2x_o`k

Explanation

To find the velocity with which the particle starts, we will follow these steps:

### Step 1: Understand the motion equation
The motion of the particle is described by the equation:
\[ x(t) = x_0(1 - e^{-kt}) \]
where \( x_0 > 0 \) and \( k > 0 \).

### Step 2: Differentiate the position function to find velocity
To find the velocity \( v(t) \), we need to differentiate the position function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left[x_0(1 - e^{-kt})\right] \]

### Step 3: Apply the derivative
Using the constant multiple rule and the chain rule, we differentiate:
1. The derivative of \( x_0 \) (a constant) is 0.
2. The derivative of \( -e^{-kt} \) is \( -(-k)e^{-kt} = ke^{-kt} \).

Thus, we have:
\[ v(t) = x_0 \cdot k \cdot e^{-kt} \]

### Step 4: Evaluate the velocity at the start (t = 0)
Now, we need to find the velocity at the start, which corresponds to \( t = 0 \):
\[ v(0) = x_0 \cdot k \cdot e^{-k \cdot 0} \]
Since \( e^{0} = 1 \), this simplifies to:
\[ v(0) = x_0 \cdot k \]

### Conclusion
The velocity with which the particle starts is:
\[ v(0) = x_0 \cdot k \]