- #1

- 178

- 0

they are looking for average speed to grannys house and then from grannys house.

can someone give me a push in the right direction, im sufficiently lost.

- Thread starter oldunion
- Start date

- #1

- 178

- 0

they are looking for average speed to grannys house and then from grannys house.

can someone give me a push in the right direction, im sufficiently lost.

- #2

cepheid

Staff Emeritus

Science Advisor

Gold Member

- 5,192

- 36

Do this again for the way back.

- #3

- 178

- 0

well for the way back i got 63 as the correct answer through guessing. 63 was nearly correct for the way there.cepheid said:

Do this again for the way back.

well in my head i can see that the first half will be 30 minutes. by calculate im lost. im still not understanding much of the equations in this physics book, or if there are physics equations?

- #4

- 178

- 0

just got answer of 60, dont ask how.

- #5

cepheid

Staff Emeritus

Science Advisor

Gold Member

- 5,192

- 36

It was a lucky guess. You assumed that the average speed would be the average of the two speeds given, which is true only because she drove for the same amount of time at each speed. The two velocities therefore had the same *weighting* in the average.oldunion said:well for the way back i got 63 as the correct answer through guessing. 63 was nearly correct for the way there.

The average of the two velocities: (50 + 76) / 2 = 126/2 = 63 mph

Average velocity according to the definition distance / time:

= [50(t1) + 76(t2)] / t_total

The problem tells us that t1 = t2 = t (she drives at each spped for the same amount of time). So obviously t_total = 2t So the equation becomes:

[50(t1) + 76(t2)] / 2t = 126t / 2t = 63

In this case, the two methods give the same result. But what if she had not driven at each speed for the same amount of time? For instance, if she drove at 76 mph for LONGER than she drove for 50 mph, obviously the average speed would be higher than 63, and would be closer to 76 than to 50. 76 would have a higher *weighting* in the average. If she had driven for longer at 50mph than at 76mph, then her average speed would have been lower, closer to 50 than to 76. This is the case on the way TO Granny's house. She does not drive at each speed for the same amount of time. t1 is not equal to t2. Use the method I told you about in my first post to find each of these times, and solve for the average speed.

Huh? Are you sure? Are you talking about on the way there? Because if she drives halfway (50 miles) at 50 miles PER HOUR, then it's obviously going to take her a full hour to do so.oldunion said:well in my head i can see that the first half will be 30 minutes.

Last edited:

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 819

- Last Post

- Replies
- 9

- Views
- 1K

- Last Post

- Replies
- 11

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 20K

- Last Post

- Replies
- 9

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 555

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 823

- Last Post

- Replies
- 3

- Views
- 1K