Could someone please review my answers and tell me if I am on the right track? Thank you!
A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0o south of
east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?
a.) average speed: 12 km/18 min=0.67 km/min
b.) average velocity: (must find time first)
t: 10.3 km/ 0.67 km = 15.4 min
average velocity: 10.3 km/15.4 min = 0.67 km/min
c.) average speed: 24 km/450 min = 0.05 km/min
average velocity: 0 km/450 min = 0
(a) we usually express car speeds in km/hour
(b) Displacement = 10.3 km in direction 25° S of east
Time = 18 min = 0.3 hour
So velocity = 34.3 km/hr in direction 25° S of E.
(c)Correct.
Thank you very much!
I wondered if I should switch to km/hr. And, I was SURE I was wrong on B but AMAZED I was right on C. :)
You're welcome!
Yes, velocity is a vector, so it is described by the magnitude (speed) and direction.
The same goes for displacement, which is also a vector, and therefore is described by the magnitude and direction.
Keep up the good work!
so i got all answers but part c for average velocity is zero since there is no info about the displacement?
For part c, you put average speed as 0.05 km/min, which is correct.
Subsequently, as part 2 of (c), you did put average velocity as 0.
Perhaps there was a confusion of the questions.
I am confused on how you got the 24km in part c. Could someone please explain that?
To review your answers, let's break down each part and see if you're on the right track.
a) To calculate average speed, we need to divide the distance traveled by the time taken. You correctly calculated it as 12 km divided by 18 min, which gives you an average speed of 0.67 km/min. Great job!
b) To find average velocity, we need to consider both the magnitude and direction of the displacement. In this case, the displacement is given as 10.3 km in a direction 25.0° south of east. However, you made a mistake in calculating the time taken. The correct calculation is to divide the distance (10.3 km) by the speed (0.67 km/min), resulting in a time of approximately 15.37 min. Notice that the direction is not relevant for calculating the average velocity since it is given as a straight-line distance. Therefore, your average velocity is indeed 0.67 km/min. Good effort!
c) To calculate the average speed and velocity for the entire trip, we need to consider both the outbound and return journeys. You correctly calculated the time for the return journey as 7 hours 30 minutes, which is equivalent to 450 minutes. However, you made a mistake in calculating the average speed. The total distance traveled for the entire trip is 2 times the initial distance, which is 24 km (12 km outbound and 12 km return). Dividing this by the total time taken (18 min outbound + 450 min return), we get an average speed of approximately 0.05 km/min.
Regarding average velocity, since the trip starts and ends at the same location, the displacement is zero. This means the average velocity is also zero, as you correctly stated.
In summary, you were on the right track with most of your answers. Just double-check the calculations for time and average speed in parts b and c. Keep up the good work!