**Given Data:**
Initial velocity (u) = 0 (as the body starts from rest)
Final velocity (v) = 72 km/hr = 72 × (5/18) m/s = 20 m/s
Time taken (t) = 20 seconds
Distance covered (s) = 80 m

**Calculating Acceleration:**
Acceleration (a) can be calculated using the equation:
v = u + at

Here, u = 0 (initial velocity), v = 20 m/s (final velocity), and t = 20 seconds.
Substituting the values into the equation, we get:
20 = 0 + a × 20
20 = 20a
a = 20/20
a = 1 m/s²

Therefore, the acceleration of the body is 1 m/s².

**Calculating Retardation:**
Retardation is the negative acceleration, denoted by (-a). It can be calculated using the equation:
v² = u² + 2as

Here, u = 20 m/s (initial velocity), v = 0 (final velocity), and s = 80 m (distance covered).
Substituting the values into the equation, we get:
0 = (20)² + 2 × (-a) × 80
0 = 400 - 160a
160a = 400
a = 400/160
a = 2.5 m/s²

Therefore, the retardation of the body is 2.5 m/s².

**Explanation:**
The given problem involves two different scenarios: the body's acceleration from rest to a certain speed and its subsequent retardation to come to rest.

In the first scenario, the body starts from rest, and its acceleration is determined using the equation v = u + at. By substituting the given values into the equation, we find the acceleration to be 1 m/s².

In the second scenario, the body decelerates to come to rest after covering a certain distance. We calculate the retardation using the equation v² = u² + 2as. By substituting the given values into the equation, we find the retardation to be 2.5 m/s².

Hence, the body experiences an acceleration of 1 m/s² in the initial phase and a retardation of 2.5 m/s² in the later phase.

the correct answer is :a = 1 m/s^2and retardation = -2.5 m/S^2