Analysis:

To solve this problem, we need to consider the relative motion of the man in the car and his friend on the motorcycle. Let's break down the problem into different stages:

Stage 1: Initial conditions
- The man in the car starts at the same time as his friend on the motorcycle, but at a distance of 100m.
- The car has a maximum constant speed of 20m/s.

Stage 2: Friend's motion
- The friend on the motorcycle starts with an initial velocity of 0m/s and undergoes constant acceleration, denoted as 'a'.
- We can use the equation of motion for uniformly accelerated motion:
- s = ut + (1/2)at^2
- where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
- Since the friend starts from rest, the equation simplifies to:
- s = (1/2)at^2

Stage 3: Man's motion
- The man in the car is moving at a constant speed of 20m/s.
- The time taken for the man to cover a distance of 100m is given by:
- t = s/v
- where s is the distance and v is the velocity.
- Substituting the values, we get:
- t = 100/20 = 5s

Stage 4: Meeting point
- To find the meeting point, we need to equate the distances covered by the friend and the man.
- The distance covered by the friend is given by:
- s_friend = (1/2)at^2
- The distance covered by the man is given by:
- s_man = vt
- Equating the two distances, we get:
- (1/2)at^2 = vt
- Simplifying the equation, we have:
- (1/2)at^2 - vt = 0
- Factoring out 't', we get:
- t((1/2)at - v) = 0
- Since time cannot be zero, we have:
- (1/2)at - v = 0
- Solving for 'a', we get:
- a = 2v/t

Final Answer:

- Substituting the given values, we have:
- a = 2(20)/5 = 8m/s^2
- Therefore, the acceleration of the friend's motorcycle must be 8m/s^2 for the man in the car to reach his friend at a distance of 100m.

Conclusion:

The correct value for the acceleration 'a' is 8m/s^2, which falls within the given range of '0,2'. This means that for any acceleration value between 0 and 2m/s^2, the man in the car will reach his friend on the motorcycle at a distance of 100m.