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 constant. How much distance is covered by the particle in returning to the starting point?

Options

A. `(2r_(0))/(a)`
B. `(r_(0))/(a)`
C. `(r_(0))/(2a)`
D. `(r_(0))/(4a)`

Explanation

To solve the problem, we need to find the total distance covered by the particle when it returns to the starting point. The position vector of the particle is given by:

\[ \vec{r}(t) = \vec{r}_0 (1 - at)t \]

where \( t \) is the time, \( a \) is a constant, and \( \vec{r}_0 \) is also a constant vector.

### Step 1: Find the velocity of the particle
The velocity \( \vec{v}(t) \) is the derivative of the position vector with respect to time \( t \):

\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \vec{r}_0 \frac{d}{dt}[(1 - at)t] \]

Using the product rule of differentiation:

\[ \vec{v}(t) = \vec{r}_0 \left(1 - at + t \frac{d}{dt}(-at)\right) = \vec{r}_0 \left(1 - at - at\right) = \vec{r}_0 (1 - 2at) \]

### Step 2: Determine when the particle changes direction
The particle changes direction when the velocity is zero:

\[ \vec{v}(t) = 0 \implies 1 - 2at = 0 \]

Solving for \( t \):

\[ 2at = 1 \implies t = \frac{1}{2a} \]

### Step 3: Calculate the position at \( t = \frac{1}{2a} \)
Now we substitute \( t = \frac{1}{2a} \) back into the position vector to find the distance \( x \):

\[ \vec{r}\left(\frac{1}{2a}\right) = \vec{r}_0 \left(1 - a\left(\frac{1}{2a}\right)\left(\frac{1}{2a}\right)\right) \]

Calculating the expression inside the parentheses:

\[ = \vec{r}_0 \left(1 - \frac{1}{4}\right) = \vec{r}_0 \left(\frac{3}{4}\right) \]

### Step 4: Find the distance \( x \)
The distance \( x \) from the starting point is the magnitude of the position vector:

\[ x = \left|\vec{r}_0 \left(\frac{3}{4}\right)\right| = \frac{3}{4} |\vec{r}_0| \]

### Step 5: Calculate the total distance covered
The total distance covered by the particle when it returns to the starting point is twice the distance \( x \):

\[ \text{Total distance} = 2x = 2 \times \frac{3}{4} |\vec{r}_0| = \frac{3}{2} |\vec{r}_0| \]

### Final Answer
Thus, the total distance covered by the particle in returning to the starting point is:

\[ \frac{3}{2} |\vec{r}_0| \]

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