\[ \frac{R}{H}=\frac{4}{tan\theta} \]

\[ As, R=H; tan\theta=4 \]

The numerical ratio of distance to the magnitude of displacement depends on the type of motion:

1. For straight-line motion in one direction, the distance and displacement are the same, so the ratio is:
\[
\frac{\text{Distance}}{\text{Displacement}} = 1
\]

2. For any other type of motion (like a curved path or circular motion), the distance is generally greater than or equal to the displacement, making the ratio:
\[
\frac{\text{Distance}}{\text{Displacement}} \geq 1
\]
The ratio is greater than 1 because distance is the total path travelled, while displacement is the shortest straight line between the start and end points.

Displacement = 2R = 80 m

Time = 4o sec

Velocity= 80 m/ 40 sec = 2 m/s

\[ t =\frac{2u}{g}=\frac{2\times 30}{10}= 6 sec\]

The stopping distance is proportional to the square of the speed:
\[
\frac{s_2}{s_1} = \left( \frac{v_2}{v_1} \right)^2
\]

Substituting the values:
\[
\frac{s_2}{6} = \left( \frac{100}{50} \right)^2 = 2^2 = 4
\]
\[
s_2 = 6 \times 4 = 24 \, \text{m}
\]

Thus, the minimum stopping distance at 100 km/h is 24 meters.

Slope of x-t graph represents velocity. Slope is negative at point E so, velocity is negative for E.

1. Distance covered: \(d = 5 \, \text{m}\) (along the semicircular path).

2. Displacement: The displacement is the straight-line distance from the starting point to the endpoint. For a semicircle with a radius \(r\):

\[
\text{Diameter} = 2r
\]
Since the distance covered is the semicircle's arc length:
\[
\text{Arc length} = \frac{1}{2}(2\pi r) = \pi r
\]

Therefore, if \(d = 5\):
\[
r = \frac{5}{\pi}
\]
So, the displacement (which is the diameter) is:
\[
\text{Displacement} = 2r = 2 \cdot \frac{5}{\pi} = \frac{10}{\pi} \, \text{m}
\]

3. Ratio of distance to displacement:
\[
\text{Ratio} = \frac{d}{\text{Displacement}} = \frac{5}{\frac{10}{\pi}} = \frac{5 \pi}{10} = \frac{\pi}{2}
\]

The resultant velocity of the swimmer is the vector sum of the river's velocity and the swimmer's velocity relative to the water.

Given:
- River velocity = 6 km/h
- Swimmer's velocity relative to water = 9 km/h

Using the Pythagorean theorem:

\[
v_{\text{resultant}} = \sqrt{(9^2 + 6^2)} = \sqrt{81 + 36} = \sqrt{117}
\]

Thus, the resultant velocity of the swimmer is:

\[
{\sqrt{117} \text{km/h}}
\]

To find the velocity of the monkey as observed by a man standing on the ground, we need to add the velocity of the monkey relative to the train to the velocity of the train.

 Given:
- Velocity of the train (north direction) = 54 km/h
- Velocity of the monkey relative to the train (opposite to train's motion) = 18 km/h

Monkey's velocity relative to the ground:
Since the monkey is running against the train’s motion, the monkey's velocity relative to the ground will be:

\[
\vec{v}_{\text{monkey}} = \vec{v}_{\text{train}} - \vec{v}_{\text{monkey relative to train}}
\]

\[
v_{\text{monkey}} = 54 \, \text{km/h} - 18 \, \text{km/h} = 36 \, \text{km/h}
\]

Thus, the velocity of the monkey as observed by a man on the ground is:

\[
{36 \, \text{km/h} \text{ north} or 10 m/s \text{ north}}
\]

The relative velocity of the boat with respect to the water is given by subtracting the velocity of the water from the velocity of the boat.

\[
\vec{v}_{\text{bw}} = \vec{v}_{\text{boat}} - \vec{v}_{\text{water}}
\]

Given:
\[
\vec{v}_{\text{boat}} = 3\hat{i} + 4\hat{j}, \quad \vec{v}_{\text{water}} = -3\hat{i} - 4\hat{j}
\]

Now, subtract:

\[
\vec{v}_{\text{bw}} = (3\hat{i} + 4\hat{j}) - (-3\hat{i} - 4\hat{j})
\]
\[
\vec{v}_{\text{bw}} = 3\hat{i} + 4\hat{j} + 3\hat{i} + 4\hat{j}
\]
\[
\vec{v}_{\text{bw}} = 6\hat{i} + 8\hat{j}
\]

Thus, the relative velocity of the boat with respect to the water is:

\[ {6\hat{i} + 8\hat{j}} \]