Final answer:
At t = 0, the car is 2.0 km west of the light, and at t = 6.0 min, the car is 5.0 km east of the light. The car's displacement between 0 min and 6.0 min is 7.0 km eastward.
At t = 3.0 min, the car's position would be 3.51 km eastward from its initial position. The car's speed at t = 6.0 min is 0.83 km/min.
Step-by-step explanation:
(a) The car's position vectors at t = 0 and t = 6.0 min can be determined by considering the distances in the positive x direction (eastward) and the negative x direction (westward). At t = 0, the car is 2.0 km west of the light, so its position vector is -2.0 km. At t = 6.0 min, the car is 5.0 km east of the light, so its position vector is 5.0 km.
(b) The car's displacement between 0 min and 6.0 min can be found by subtracting the initial position vector from the final position vector. The displacement is 5.0 km - (-2.0 km) = 7.0 km in the positive x direction (eastward).
(c) To find the car's position at t = 3.0 min, we can use the concept of constant velocity. Since the car travels a total distance of 7.0 km in 6.0 min, its average velocity is 7.0 km / 6.0 min = 1.17 km/min. Therefore, at t = 3.0 min, the car would have traveled (1.17 km/min) * (3.0 min) = 3.51 km eastward from its initial position, placing it at a position vector of 3.51 km.
(d) The car's speed at t = 6.0 min can be determined by dividing the distance traveled by the time taken. The car travels a total distance of 5.0 km in 6.0 min, so its speed is 5.0 km / 6.0 min = 0.83 km/min.