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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 $25.0^\circ$ 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?

Question by OpenStax is licensed under CC BY 4.0.

a) $40.0 \textrm{ km/h}$

b) $34.3 \textrm{ km/h}$

c) average speed = $3.20 \textrm{ km/h}$, average velocity = $0$

Solution Video

OpenStax College Physics Solution, Chapter 2, Problem 11 (Problems & Exercises) (2:57)

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Video Transcript
This is College Physics Answers with Shaun Dychko I've drawn the student's home here and her university over here. We know that the straight line between her home and the university is at an angle of 25.0 degrees towards the South compared to east or south of east. And it's 10.3 kilometers long, but she's not going to drive along that straight line path. Instead, she's going to take this green path and have a total distance travelled of 12.0 kilometers when she goes from her home to the university. And it takes her 18.0 minutes to do this trip and we'll convert that into hours because our speeds and velocities should be expressed in units of kilometers per hour. So, we multiply 18.0 minutes by 1 hour for every 60 minutes and that gives us .3 hours. So, the average speed of the trip from our home to the university is the distance traveled divided by the time. So, that's 12 kilometers divided by 0.3 hours, which gives 40.0 kilometers per hour. The average velocity on the other hand is the displacement or the change in position and you don't think about the path that she took, but instead, think about just what is the final position compared to the initial position. And those final and initial positions are separated by 10.3 kilometres. And we divide that by, this time, .3 hours and that gives 34.3 kilometers per hour average velocity. And then, if she returns home later and she arrives home 7 hours and 30 minutes after she left her home. So, that means the total time that she spent travelling is 7.5 hours. Or I just... You know, I put .5 there because 30 minutes is half an hour. And we can all assume that the 30 minutes has 2 significant figures in which case we can put 7.50 hours. So, put 2 significant figures in the decimal part there. And then, we have 2 times the distance traveled. It is the total distance she goes because she goes 12 kilometers in the way there and then follows the same way in the 12 kilometer path back to her home. And so, this is 2 times 12 kilometers divided by 7.5 hours, which works out to 3.2 kilometers per hour is her average speed overall for the entire day. The average velocity is 0 because her final position and her initial position are separated by 0 meters because she arrives at the place where she started. And so, there's no displacement. And so, therefore no velocity either.


Submitted by lesiramdelgado on Fri, 09/04/2020 - 16:50

quick question why is 7.50 since is half later would it be 7.30 im confuse

Submitted by vicente_cabanillas on Sat, 09/05/2020 - 21:53

7 hours and 30 minutes is 7 1/2 hours. Half of one hour is 30s or 1/2 hr. The question is being asked and answered in hours. 7 whole hours is 7.0 half of an hour is .5 since you are dividing one in half (not in seconds) so it is 7.50 hrs (not 7hours and 30 seconds). remember 1/2 is 0.5
So if half an hour is 0.50hr and you add 0.50 to 7.50 you get 8.0hrs = 8 whole hours

In reply to by lesiramdelgado

Submitted by gaigewilder on Thu, 01/21/2021 - 12:34

would it not actually be 7.8hrs? the initial trip took 18.0 mins. if you are counting the total time of the trip we should be including that.


24.0km/7.8h =approx 3.1km/h