The position vector of a particle is given by `vecr_()` = `vecr_(0)` (1-at)t, where t is the time and a as well as `vecr_(0)` are constantwhat will be the velocity of the particle when it returns to the starting point?

Options

A. `vecr_(0)`
B. `-vecr_(0)`
C. `2vecr_(0)`
D. `-2vecr_(0)`

Explanation

To solve the problem step by step, we will analyze the given position vector of the particle and determine its velocity when it returns to the starting point.

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = \vec{r}_0 (1 - at)t \]
where \( \vec{r}_0 \) is a constant vector, \( a \) is a constant, and \( t \) is the time.

### Step 2: Find the time when the particle returns to the starting point
To find when the particle returns to the starting point, we set the position vector equal to zero:
\[ \vec{r}(t) = 0 \]
This gives us:
\[ \vec{r}_0 (1 - at)t = 0 \]
Since \( \vec{r}_0 \) is not zero, we can focus on the scalar part:
\[ (1 - at)t = 0 \]
This equation holds true if either \( t = 0 \) or \( 1 - at = 0 \). Solving for \( t \):
1. \( t = 0 \) (the initial position)
2. \( 1 - at = 0 \) leads to \( at = 1 \) or \( t = \frac{1}{a} \)

Thus, the particle returns to the starting point at \( t = 0 \) and again at \( t = \frac{1}{a} \).

### Step 3: Differentiate the position vector to find the velocity vector
Next, we differentiate the position vector with respect to time \( t \) to find the velocity vector:
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} \]
Substituting the expression for \( \vec{r}(t) \):
\[ \vec{v}(t) = \frac{d}{dt} \left( \vec{r}_0 (1 - at)t \right) \]
Using the product rule:
\[ \vec{v}(t) = \vec{r}_0 \left( \frac{d}{dt}(1 - at) \cdot t + (1 - at) \cdot \frac{d}{dt}(t) \right) \]
Calculating the derivatives:
\[ \frac{d}{dt}(1 - at) = -a \quad \text{and} \quad \frac{d}{dt}(t) = 1 \]
Thus, we have:
\[ \vec{v}(t) = \vec{r}_0 \left( -a t + (1 - at) \cdot 1 \right) \]
Simplifying this:
\[ \vec{v}(t) = \vec{r}_0 (1 - 2at) \]

### Step 4: Evaluate the velocity at \( t = \frac{1}{a} \)
Now, we substitute \( t = \frac{1}{a} \) into the velocity equation:
\[ \vec{v}\left(\frac{1}{a}\right) = \vec{r}_0 \left(1 - 2a \cdot \frac{1}{a}\right) \]
This simplifies to:
\[ \vec{v}\left(\frac{1}{a}\right) = \vec{r}_0 (1 - 2) = -\vec{r}_0 \]

### Final Answer
Thus, the velocity of the particle when it returns to the starting point is:
\[ \vec{v} = -\vec{r}_0 \]