U2.8 Driving like a car

The bay is to the robot's right, between two parked cars. All the distances are from the robot's starting point, facing along +y (heading 0). - The bay is 18 cm wide and 47 cm long. Across, it runs from 13 cm to 31 cm to your right; along, from 10 cm behind you to 37 cm ahead. Its middle is 22 cm to your right and 13.5 cm ahead. - The parked cars are 14 cm wide and 22 cm long, from 16 cm to 30 cm to your right. The one behind runs from 35 cm to 13 cm behind you; the one ahead from 42 cm to 64 cm ahead. - The robot is 7 cm across, and touching a car counts from its edge, not its centre. To pass, the robot's centre must finish inside the bay, facing within 15 degrees of the way it started (a final heading between 345 and 15), without touching either car, inside 40 seconds. Every move must go through car(), kept exactly as it is given with V_MAX, W_MAX and R_MIN unchanged: no other drive(), no sliding and no turning on the spot.

from bugbot import *
import math
connect()

V_MAX, W_MAX, R_MIN = 20.0, 120.0, 25.0

def car(speed, steer, seconds):
    """speed -100..100 (negative reverses), steer -1..1 (full lock left..right). Turn rate is tied to speed."""
    speed = max(-100, min(100, speed))
    steer = max(-1, min(1, steer))
    v = speed / 100 * V_MAX
    w = math.degrees(v / R_MIN) * steer
    drive(speed, 0, 100 * w / W_MAX)
    wait(seconds)
    stop()
    wait(0.4)

# your moves, all with car()

The hint students can ask for: Work out, from the numbers in the task, how far sideways and how far along the bay's middle is from where you start. The two-arc formulas turn the sideways distance into the angle each arc must turn, and that angle into the length of road the arcs use; so you know how far forward to go before reversing. Then turn each distance into seconds from the car's speed. Check the forward leg keeps you clear of the car ahead.

A solution

from bugbot import *
import math
connect()

V_MAX, W_MAX, R_MIN = 20.0, 120.0, 25.0

def car(speed, steer, seconds):
    """speed -100..100 (negative reverses), steer -1..1 (full lock left..right). Turn rate is tied to speed."""
    speed = max(-100, min(100, speed))
    steer = max(-1, min(1, steer))
    v = speed / 100 * V_MAX
    w = math.degrees(v / R_MIN) * steer
    drive(speed, 0, 100 * w / W_MAX)
    wait(seconds)
    stop()
    wait(0.4)

# aim 22 cm to the right and 11 cm ahead: the middle of the bay across, a little behind its middle along (away from the car ahead)
SIDE, AHEAD = 22, 11
v = 60 / 100 * V_MAX                                # cm/s at 60 percent
phi = math.acos(1 - SIDE / (2 * R_MIN))             # each arc turns this far: 2 R (1 - cos phi) = SIDE
back = 2 * R_MIN * math.sin(phi)                    # and the two arcs together take this much road
arc = R_MIN * phi / v                               # seconds for one arc

car(60, 0, (AHEAD + back) / v)                      # past the bay, far enough to reverse into it
car(-60, 1, arc)                                    # swing the back in
car(-60, -1, arc)                                   # straighten up
print("phi", round(math.degrees(phi)), "arc", round(arc, 2), "s, then", round((AHEAD + back) / v, 2), "s")

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.