We can use the work-energy principle to find the force.

Given:
- Initial velocity, \( u = 20 \, \text{m/s} \)
- Final velocity, \( v = 5 \, \text{m/s} \)
- Distance, \( s = 100 \, \text{m} \)
- Mass, \( m = 20 \, \text{kg} \)

Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
\[
(5)^2 = (20)^2 + 2a(100)
\]
\[
25 = 400 + 200a
\]
\[
200a = -375 ;a = -\frac{375}{200} = -1.875 \, \text{m/s}^2
\]

Now, force \( F = ma \):
\[
F = 20 \times (-1.875) = -37.5 \, \text{N}
\]

Thus, the force on the body is -37.5 N (opposite to the direction of motion).

Using the formula

$a = \frac{v_f - v_i}{t}$

:

Converting to m/s:

$90$

km/h =

$25$

m/s,

$54$

km/h =

$15$

m/s.

$$a = \frac{15 - 25}{15} = \frac{-10}{15} = -0.67 \text{ m/s}^2$$

Retardation = 0.67 m/s².

Using

$t = \frac{s}{v}$

, with

$v = 54$

km/h = 15 m/s and

$s = 5$

m:

$$t = \frac{5}{15} = 0.33 \text{ s}$$

Reaction time = 0.33 s.

Using the stopping distance formula:

$$s \propto u^2$$

If speed is halved, new stopping distance:

$$s_2 = \frac{100}{4} = 25 \text{ m}$$

Distance in \( n^{\text{th}} \) second is \( S_n = u + frac {a}{2}(2n-1)) \). Starting from rest \( u=0 , S_n =frac {a}{2}(2n-1).\) Given

\( S_n = 3 S_{n-1} \), we have 2n-1 = 3(2n-3). Solving gives (4n = 8), hence (n = 2).

Stopping distance \(s\) is proportional to \(u^2\). When the speed is doubled (from 50 to 100), the stopping distance increases by a factor of 4. Thus, \(s' = 4 \times 6 = 24\text{ m}\).

Max velocity \( v = 20 \text{ ms}^{-1}\). Acceleration distance \(s_1 = \frac{1}{2}(2)(10^2) = 100\text{ m}\). Constant velocity distance \(s_2 = 20 \times 30 = 600\text{ m}\). Deceleration distance \(s_3 = \frac{20^2}{2 \times 4} = 50\text{ m}\).

Total distance = 750 m.

Initial velocity along x-axis is \(v_x = 0.4\text{ m/s}\). Velocity developed along the perpendicular y-axis is \(v_y = a_y t = 0.15 \times 2 = 0.3\text{ m/s}\). Resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m/s}\).

Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).

Since acceleration is uniform, the distance is given by the formula \(s = \frac{u+v}{2} \times t\). Substituting \(u=20\text{ m/s}\), \(v=60\text{ m/s}\), and \(t=5\text{ s}\), we get \(s = \frac{20+60}{2} \times 5 = 200\text{ m}\).