A basketball player runs around the basketball court during a training session. In one of such sessions, he runs from A to D, then back to the centre line from D, goes back to D and finally back to A. Calculate the total distance covered by the player.

The court has a length of $30 \text{ m}$ (A to D or B to C) and a breadth of $15 \text{ m}$ (A to B or D to C).
The player's path is:

• From A to D: $30 \text{ m}$ (along the length).
• Then back to the centre line from D. The "centre line" (half-court line) is typically halfway along the length of the court. So, from D to the midpoint of AD is $30 \text{ m} / 2 = 15 \text{ m}$.
• Goes back to D: From the centre line back to D is $15 \text{ m}$.
• Finally back to A: From D to A is $30 \text{ m}$ (along the length).

The total distance covered by the player is $\boxed{\text{90 m}}$.

Average speed is defined as total distance divided by total time.

The time it took the player to complete the session is $\boxed{\text{36 s}}$.

The player starts at point A and finishes at point A.

Since the initial and final positions are the same, the total displacement is $0 \text{ m}$.

The average velocity of the athlete is $\boxed{\text{0 m/s}}$.

The speed of car A is $\boxed{\text{37.5 m/s}}$.

Initial velocity $u = 0 \text{ m/s}$ (starts from rest).
Acceleration $a = 1.5 \text{ m/s}^2$.
Time $t = 11.5 \text{ s}$.

The velocity of car B is $\boxed{\text{17.25 m/s}}$.

Initial velocity $u = 0 \text{ m/s}$.
Acceleration $a = 1.5 \text{ m/s}^2$.
Time $t = 11.5 \text{ s}$.

Car B has driven $\boxed{\text{99.19 m}}$ (rounded to two decimal places).

Final velocity $v = 0 \text{ m/s}$ (comes to a standstill).
Distance $s = 40 \text{ m}$.
Time $t = 4 \text{ s}$.

The acceleration car A experiences is $\boxed{\text{-5 m/s}^2}$. The negative sign indicates deceleration.