Suppose that the position function for a particle is given as `vec( r ) ( t) = x ( t) hat( i) + y (t) hat(j) with ` x ( t) = t+1` and `y ( t) = ( t^(2) //8) + 1`. The ratio of magnitude of average velocity ( during time interval t = 2 to 4 sec) to the speed at t =2 sec is

Options

A. `sqrt( 3) : 2`
B. ` sqrt( 5) : 2`
C. `sqrt( 5) : sqrt(3)`
D. none of the above

Answer

To solve the problem, we need to find the ratio of the magnitude of average velocity during the time interval from \( t = 2 \) seconds to \( t = 4 \) seconds to the speed at \( t = 2 \) seconds. We will follow these steps:

### Step 1: Define the position function
The position function is given as:
\[ \vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} \]
where
\[ x(t) = t + 1 \]
\[ y(t) = \frac{t^2}{8} + 1 \]

### Step 2: Calculate the position at \( t = 2 \) seconds and \( t = 4 \) seconds
- For \( t = 2 \):
\[ x(2) = 2 + 1 = 3 \]
\[ y(2) = \frac{2^2}{8} + 1 = \frac{4}{8} + 1 = 0.5 + 1 = 1.5 \]
Thus,
\[ \vec{r}(2) = 3 \hat{i} + 1.5 \hat{j} \]

- For \( t = 4 \):
\[ x(4) = 4 + 1 = 5 \]
\[ y(4) = \frac{4^2}{8} + 1 = \frac{16}{8} + 1 = 2 + 1 = 3 \]
Thus,
\[ \vec{r}(4) = 5 \hat{i} + 3 \hat{j} \]

### Step 3: Calculate the displacement
The displacement \( \Delta \vec{r} \) from \( t = 2 \) to \( t = 4 \) is:
\[ \Delta \vec{r} = \vec{r}(4) - \vec{r}(2) = (5 \hat{i} + 3 \hat{j}) - (3 \hat{i} + 1.5 \hat{j}) = (5 - 3) \hat{i} + (3 - 1.5) \hat{j} = 2 \hat{i} + 1.5 \hat{j} \]

### Step 4: Calculate the magnitude of the displacement
The magnitude of the displacement is:
\[ |\Delta \vec{r}| = \sqrt{(2)^2 + (1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \]

### Step 5: Calculate the average velocity
The average velocity \( \vec{V}_{avg} \) is given by:
\[ \vec{V}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{2.5}{4 - 2} = \frac{2.5}{2} = 1.25 \]

### Step 6: Calculate the speed at \( t = 2 \) seconds
To find the speed, we first need the velocity function \( \vec{v}(t) \):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(x(t) \hat{i} + y(t) \hat{j}) = \frac{d}{dt}(t + 1) \hat{i} + \frac{d}{dt}\left(\frac{t^2}{8} + 1\right) \hat{j} \]
Calculating the derivatives:
\[ \vec{v}(t) = 1 \hat{i} + \left(\frac{2t}{8}\right) \hat{j} = 1 \hat{i} + \frac{t}{4} \hat{j} \]
Now, at \( t = 2 \):
\[ \vec{v}(2) = 1 \hat{i} + \frac{2}{4} \hat{j} = 1 \hat{i} + 0.5 \hat{j} \]
The magnitude of the velocity (speed) at \( t = 2 \) is:
\[ |\vec{v}(2)| = \sqrt{(1)^2 + (0.5)^2} = \sqrt{1 + 0.25} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \]

### Step 7: Calculate the ratio of average velocity to speed
Now we can find the ratio:
\[ \text{Ratio} = \frac{|\vec{V}_{avg}|}{|\vec{v}(2)|} = \frac{1.25}{\frac{\sqrt{5}}{2}} = \frac{1.25 \times 2}{\sqrt{5}} = \frac{2.5}{\sqrt{5}} = \frac{2.5 \sqrt{5}}{5} = \frac{\sqrt{5}}{2} \]

### Final Answer
Thus, the ratio of the magnitude of average velocity to the speed at \( t = 2 \) seconds is:
\[ \frac{\sqrt{5}}{2} \]