What is the Average Speed Computation Problem on a Hill?

[Bashyboy]

Homework Statement

The problem is in the provided link (it is questions 2.2) http://www.docstoc.com/docs/3817109/Chapter-Problems-A-car-travels-up-a-hill-at-a

Homework Equations

The Attempt at a Solution

I understand the duration of time it takes to go up the hill is a larger time interval than going down; but I fail to see that fact as a viable way to justify not just simply adding the to velocities and dividing by two. Could someone please understand this part to me, thank you.

 

[PeterO]

Homework Statement

The problem is in the provided link (it is questions 2.2) http://www.docstoc.com/docs/3817109/Chapter-Problems-A-car-travels-up-a-hill-at-a

Homework Equations

The Attempt at a Solution

I understand the duration of time it takes to go up the hill is a larger time interval than going down; but I fail to see that fact as a viable way to justify not just simply adding the to velocities and dividing by two. Could someone please understand this part to me, thank you.

Average speed is total distance / total time. The length of the hill enables us to calculate the time for the trip - and has no other application.

Perhaps an example is easier explanation.

Suppose the hill is 10 km long, and you travel up at 10 km/h and down and 30 km/h

You are tending to an answer of 20 km/h for the average speed.

The total trip (up then down) is 20km. At 20 km/h that would take 1 hour

HOWEVER: If traveling at 10km/h on the way up, it takes 1 hour to get up the hill, so it is impossible to get up and down in 1 hour - so to simply add the velocities and divide by 2 doesn't work.

The simple average only applies if you travel at different speeds for equal times.

40 kph for 1 hour then 60 kph for 1 hour means an average speed of 50 kph
Note that you covered 40 km in the first hour then 60 km in the second hour - so it can't have represented a trip in opposite directions along the same piece of road.

 

[Bashyboy]

That was bloody brilliant, thank you so very much for taking your time to answer my question.