posted by Miaow on .
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.
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. :)
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?