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 initial velocity of the particle described by the equation \( x = x_0(1 - e^{-kt}) \), we will follow these steps:

### Step 1: Write down the equation of motion
The motion of the particle is given by:
\[ x = x_0(1 - e^{-kt}) \]

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

### Step 3: Apply the differentiation
The derivative of \( 1 \) is \( 0 \) and the derivative of \( -e^{-kt} \) is \( k e^{-kt} \) (using the chain rule):
\[ \frac{dx}{dt} = x_0 \cdot (0 - (-k e^{-kt})) = x_0 k e^{-kt} \]

### Step 4: Find the initial velocity
To find the initial velocity, we evaluate \( v \) at \( t = 0 \):
\[ v(0) = x_0 k e^{-k \cdot 0} = x_0 k e^{0} = x_0 k \]

### Step 5: Write the final answer
Thus, the initial velocity of the particle is:
\[ v = k x_0 \]