Easy avg velocity problem I can't figure out

  • Thread starter Thread starter weeeee123
  • Start date Start date
  • Tags Tags
    Figure Velocity
AI Thread Summary
The average velocity problem involves a trip where the first half is traveled at 10 MPH and the second half at 40 MPH. The key mistake is interpreting "first half" and "second half" as equal time intervals instead of equal distances. The correct approach is to calculate the total distance traveled and divide it by the total time taken. The average velocity is determined to be 16 MPH, not 25 MPH, due to the different distances covered at each speed. Understanding this distinction resolves the confusion.
weeeee123
Messages
3
Reaction score
0

Homework Statement


First half of a trip is @ 10 MPH
Second half of a trip is @ 40 MPH
Avg velocity?

Homework Equations


(Vi+Vf)/2 = avg velocity...?

The Attempt at a Solution


K, this is obviously a simple question but for some reason when I do it I get it wrong, and I'm 100% sure I'm right. But I'm not. WTH?

I know the answer is 16, because the answer is given to me. I don't know how to get to there though for whatever reason. What I did was basically weight each velocity by multiplying each by .5 since each speed is for half the time. I get 25, which is wrong. Then I thought about it this way. If I go 10 MPH for half an hour, that's 5 miles. Then if I go 40 MPH for half an hour, that's 20 miles. In total I went 25 miles in one hour, and my avg velocity should be 25 in that case. What am I missing?
 
Physics news on Phys.org
weeeee123 said:

Homework Statement


First half of a trip is @ 10 MPH
Second half of a trip is @ 40 MPH
Avg velocity?

Homework Equations


(Vi+Vf)/2 = avg velocity...?


The Attempt at a Solution


K, this is obviously a simple question but for some reason when I do it I get it wrong, and I'm 100% sure I'm right. But I'm not. WTH?

I know the answer is 16, because the answer is given to me. I don't know how to get to there though for whatever reason. What I did was basically weight each velocity by multiplying each by .5 since each speed is for half the time. I get 25, which is wrong. Then I thought about it this way. If I go 10 MPH for half an hour, that's 5 miles. Then if I go 40 MPH for half an hour, that's 20 miles. In total I went 25 miles in one hour, and my avg velocity should be 25 in that case. What am I missing?

The equation that you wrote for the average velocity is incorrect. Instead, the average velocity is the total distance divided by the total time. Try it that way instead.

(Don't worry -- this is a common error, and it's why they are asking you this question.)
 
I quite honestly didn't even use that equation. I basically did what you said, dividing the total distance traveled by the total time.

avg vel = (40 *.5t + 10 *.5t)/t
= t(40*.5 + 10 * .5)/t
= (40 * .5 + 10 * .5)
= 25

ummm...?
 
weeeee123 said:
I quite honestly didn't even use that equation. I basically did what you said, dividing the total distance traveled by the total time.

avg vel = (40 *.5t + 10 *.5t)/t
= t(40*.5 + 10 * .5)/t
= (40 * .5 + 10 * .5)
= 25

ummm...?

When they say "first half" and "second half" of the trip, they mean in distance, not in time...
 
berkeman said:
When they say "first half" and "second half" of the trip, they mean in distance, not in time...

Oh wow hahahahaha thank you
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...

Similar threads

Replies
9
Views
1K
Replies
2
Views
2K
Replies
7
Views
1K
Replies
11
Views
1K
Replies
17
Views
2K
Replies
10
Views
2K
Back
Top