To solve this, we'll use the equation of motion:

\[
v^2 = u^2 + 2as
\]

Where:
- \( v = 0 \) (final velocity, since the car comes to rest)
- \( u = 144 \, \text{km/h} = 40 \, \text{m/s} \) (initial velocity)
- \( s = 200 \, \text{m} \) (distance)
- \( a \) is the acceleration (which we'll solve for)

Rearranging the equation:

\[
0 = (40)^2 + 2 \cdot a \cdot 200
\]
\[
0 = 1600 + 400a
\]
\[
a = - \frac{1600}{400} = -4 \, \text{m/s}^2
\]

Now, to find the time (\( t \)):

\[
v = u + at
\]
\[
0 = 40 + (-4)t
\]
\[
t = \frac{40}{4} = 10 \, \text{seconds}
\]

Thus, it takes 10 seconds for the car to stop.

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

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.

\[ V_{av}=\frac{\int_{0}^{1}v.dt}{\int_{0}^{1}dt}= \frac{\int_{0}^{1}(t + 3).dt}{\int_{0}^{1}dt}=\frac{7}{2} m/s \]

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:
\[
\text{Ratio} = \frac{d}{\text{Displacement}} = \frac{5}{\frac{10}{\pi}} = \frac{5 \pi}{10} = \frac{\pi}{2}
\]

Change in a vector when it is rotated by an angle θ is

\[ \Delta V = 2V sin \left( \frac{\theta}{2} \right) \]

\[ \Delta V = 2\times 10 sin \left( \frac{90^{0}}{2} \right)= 20/\sqrt{2}= 10\sqrt{2}= 14.4 m/s \]

1. Time taken by each ball:
- First ball: \( t_1 = 5 \, \text{s} \)
- Second ball: \( t_2 = 3 \, \text{s} \) (dropped 2 s later)

2. Heights:
- Height of the first ball:
\[
h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \cdot 10 \cdot (5)^2 = 125 \, \text{m}
\]
- Height of the second ball:
\[
h_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \cdot 10 \cdot (3)^2 = 45 \, \text{m}
\]

3. Difference in heights:
\[
\Delta h = h_1 - h_2 = 125 - 45 = 80 \, \text{m}
\]

 The difference in initial heights is 80 meters.

To cross the river in the shortest time, the man should swim directly perpendicular to the riverbank, regardless of the river's flow. This way, his entire swimming speed is used to cross the river.

 Given:
- Speed of the river (east to west) = 5 m/min
- Speed of the swimmer in still water = 10 m/min

The swimmer should aim **directly north** to minimize the crossing time. The river's current will cause him to drift downstream, but this direction ensures the shortest crossing time.

Thus, the man should swim:

\[
{\text{directly north}}
\]

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:

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