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:

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