Final answer:
The student's average speed to the university was 40.0 km/h. The average velocity was approximately 34.3 km/h in the direction 25.0º south of east.
For the entire round trip, the average speed was 3.16 km/h, and the average velocity was 0 km/h because the displacement was zero.
Step-by-step explanation:
The question involves calculations of average speed and average velocity for a student's trip to the university and back home.
To calculate the average speed, we divide the total distance traveled by the total time taken. The student travelled 12.0 km in 18.0 minutes, which is 0.3 hours (18 divided by 60).
Therefore, the average speed is 12.0 km ÷ 0.3 h = 40.0 km/h.
The average velocity on the way to the university is given by the displacement divided by the time taken. Displacement is the straight-line distance in the specific direction of 25.0º south of east, which is 10.3 km.
Converting again the time to hours, we get an average velocity of 10.3 km ÷ 0.3 h = 34.333 km/h, which when rounded is approximately 34.3 km/h. The direction is not factored in the magnitude but is understood to be part of the velocity vector.
The total round trip distance is 12.0 km to the university plus 12.0 km returning home, which equals 24.0 km. The student took 18 minutes to university and 18 minutes back, plus the time spent at the university, for a total time of 7 hours and 36 minutes, or 7.6 hours.
The average speed for the round trip is 24.0 km ÷ 7.6 h = 3.158 km/h, rounded to 3.16 km/h.
Because the trip started and ended at the same point, the displacement is zero, thus the average velocity for the entire trip is 0 km/h.