Extremely annoying question, average velocity related

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SUMMARY

The discussion centers on calculating the average velocity of a truck that travels 12 km east in 20 minutes and then 4 km north in 10 minutes. The correct average velocity is determined to be 24 km/hr east and 8 km/hr north, with conversions yielding 6.7 m/s east and 2.2 m/s north. The confusion arises from misapplying the average velocity formula and incorrectly combining distances instead of calculating the resultant displacement. The participant ultimately clarifies their understanding by correctly summing the time and applying the average velocity formula accurately.

PREREQUISITES
  • Understanding of average velocity and displacement concepts
  • Ability to convert time from minutes to hours
  • Familiarity with basic trigonometry for resultant vectors
  • Knowledge of unit conversions between kilometers per hour and meters per second
NEXT STEPS
  • Study vector addition and how to resolve vectors into components
  • Learn about the Pythagorean theorem in the context of displacement
  • Practice converting units between km/hr and m/s
  • Review average velocity calculations with multiple segments of travel
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Students studying physics, particularly those learning about motion and velocity calculations, as well as educators looking for examples of common misconceptions in average velocity problems.

dejan
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Homework Statement


Ok so i have this simple question on a physics practice sheet, and because i don't get how the professor got his answers (answer are given) i have become confused and created this question into a difficult one. I am very annoyed when i can't get the exact same answer as the professors answers, and i have followed every step. Here is the question

A truck drivers 12km east in 20 minutes and then drives 4 km north in 10 minutes. What is the average velocity of the truck for the entire trip?

The Attempt at a Solution


So i draw it all...in the notes, it basically says, average velocity = (delta)x/time
so i get my delta x by subtracting the 12km-4km = 8km (i'm not going to convert here). It says average velocity for the entire trip so i gather you subtract/add the numbers...and also convert the 20minutes and 10minutes to hour seeing how you're working with km here...30min=0.5hr.
Now i put that all into the equation, av.= 8/0.5 = 16??

But the answers say, 24km/hr east and 8km/hr north (6.7m/s east, 2.2m/s north) ??

I know how those numbers came about...12km+12km=24, 4km+4km=18 and the m/s ones came from the conversions of 12km=3.33m/s and 4km=1.11m/s so you just double those...i know I'm doing something wrong here, I've made it seem too difficult when it's only supposed to be a simple exercise...

And also, the question after that one, is a bit similar, except that there is no addition or subtraction of any numbers, it just basically gives you the over all x and time which you divide both and i get the same answer...i'm going to go back and read over my notes because i think I've missed something...any help would be great thanks!
 
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dejan said:

Homework Statement


Ok so i have this simple question on a physics practice sheet, and because i don't get how the professor got his answers (answer are given) i have become confused and created this question into a difficult one. I am very annoyed when i can't get the exact same answer as the professors answers, and i have followed every step. Here is the question

A truck drivers 12km east in 20 minutes and then drives 4 km north in 10 minutes. What is the average velocity of the truck for the entire trip?

The Attempt at a Solution


So i draw it all...in the notes, it basically says, average velocity = (delta)x/time
so i get my delta x by subtracting the 12km-4km = 8km (i'm not going to convert here). It says average velocity for the entire trip so i gather you subtract/add the numbers...and also convert the 20minutes and 10minutes to hour seeing how you're working with km here...30min=0.5hr.
Now i put that all into the equation, av.= 8/0.5 = 16??

If you draw the diagram, you will see that the displacement is the hypotenuse of the triangle with sides 12 and 4.

But the answers say, 24km/hr east and 8km/hr north (6.7m/s east, 2.2m/s north) ??

I know how those numbers came about...12km+12km=24, 4km+4km=18 and the m/s ones came from the conversions of 12km=3.33m/s and 4km=1.11m/s so you just double those...i know I'm doing something wrong here, I've made it seem too difficult when it's only supposed to be a simple exercise...
Here, you've just tried to put together numbers to get the answer given on the solution sheet! This is not the way to go about problems. Adding together two distances, namely 12km+12km will not give a velocity of 24km/h!

Start by obtaining the magnitude of the resultant velocity, then find the angle in order to resolve into east and north components.

And also, the question after that one, is a bit similar, except that there is no addition or subtraction of any numbers, it just basically gives you the over all x and time which you divide both and i get the same answer...i'm going to go back and read over my notes because i think I've missed something...any help would be great thanks!

You'll have to post some details in order to be helped/
 
Wait i think i finally got it...jeez i don't know why i make things like this so hard...the times, 20min and 10min, well i added them to get 30min and converted it to km/hr so i got 0.5...and that would end up being the time right? coz if you divide 12 by 0.5 you get 24 and 4 by 0.5 is 8, which is what the answers say...so have i done this properly?
And also if u divide 3.33m/s by 0.5 you get 6.6 which is what the answers were for the m/s one...but I'm not sure if it's right to divide by 0.5 because that's in km/hr??
 

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