# Vectors: Displacement, average velocity & speed

• AidenPhysica
In summary: Thank you.In summary, a plane flies 413 km east from city A to city B in 49.0 min and then 814 km south from city B to city C in 1.70 h. The magnitude of the plane's displacement is 912.8 km and its direction is -63.098 degrees, measured from the +x direction (east). The magnitude of its average velocity is 362.2222 km/h and its direction is also -63.098 degrees. The average speed of the plane is 486.9 km/h.
AidenPhysica

## Homework Statement

A plane flies 413 km east from city A to city B in 49.0 min and then 814 km south from city B to city C in 1.70 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

## Homework Equations

so velocity is = to change in displacement over change in time

## The Attempt at a Solution

For a, the magnitude of displacement is (413km^2+814^2)^.5 pythagorean theorem. is 912.8 km right?
For b, the direction of the plane's displacement is tan^-1 (814.2/413). but this angle 63.104 must be expressed of magnitude less than 180 degrees measured from +x direction (east). So how would you do that? Because 63.104 degrees is positive and not right, do you just minus 180 degrees? Then is it just -116.902 degrees? Yeah I guess so.
For c, the magnitude of average velocity is displacement 912.8km/ 2.52 hours to get 362.2222 km/hour.
For d, is the direction of avg velocity just the same as the answer to b, is it just -116.902 degrees also?
For e, isn't avg. speed just 486.9 km/hour?
Basically I have got answers but am really unsure if I am doing it right or wrong. For all I know, I am completely missing the point. Thanks again.

AidenPhysica said:

## Homework Statement

A plane flies 413 km east from city A to city B in 49.0 min and then 814 km south from city B to city C in 1.70 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

## Homework Equations

so velocity is = to change in displacement over change in time

## The Attempt at a Solution

For a, the magnitude of displacement is (413km^2+814^2)^.5 pythagorean theorem. is 912.8 km right?
For b, the direction of the plane's displacement is tan^-1 (814.2/413). but this angle 63.104 must be expressed of magnitude less than 180 degrees measured from +x direction (east). So how would you do that? Because 63.104 degrees is positive and not right, do you just minus 180 degrees? Then is it just -116.902 degrees? Yeah I guess so.
For c, the magnitude of average velocity is displacement 912.8km/ 2.52 hours to get 362.2222 km/hour.
For d, is the direction of avg velocity just the same as the answer to b, is it just -116.902 degrees also?
For e, isn't avg. speed just 486.9 km/hour?
Basically I have got answers but am really unsure if I am doing it right or wrong. For all I know, I am completely missing the point. Thanks again. [/B]
Is the answer to d and b the same?

AidenPhysica said:
Because 63.104 degrees is positive and not right
I think it is common for angles to be considered positive as you rotate CCW from the +x axis. So I would think that your angle would be -63.1 degrees. But it all depends on who defined it.

I think your solution looks correct.

AidenPhysica said:

## The Attempt at a Solution

For a, the magnitude of displacement is (413km^2+814^2)^.5 pythagorean theorem. is 912.8 km right?
For b, the direction of the plane's displacement is tan^-1 (814.2/413). but this angle 63.104 must be expressed of magnitude less than 180 degrees measured from +x direction (east). So how would you do that? Because 63.104 degrees is positive and not right, do you just minus 180 degrees? Then is it just -116.902 degrees? Yeah I guess so.
Subtracting 180 degrees would reverse the direction of the vector. You can only add or subtract multiples of 360° if you want to retain the direction.

In this case you've lost the sign of the north-south displacement. When the plane traveled southwards from B, what sign should you assign the displacement?
For c, the magnitude of average velocity is displacement 912.8km/ 2.52 hours to get 362.2222 km/hour.
For d, is the direction of avg velocity just the same as the answer to b, is it just -116.902 degrees also?
For e, isn't avg. speed just 486.9 km/hour?
Basically I have got answers but am really unsure if I am doing it right or wrong. For all I know, I am completely missing the point. Thanks again.
Your answer to (a) is good. Fix up the answer to (b) by setting the correct sign on the displacements. You've got the right ideas for the rest.

gneill said:
Subtracting 180 degrees would reverse the direction of the vector. You can only add or subtract multiples of 360° if you want to retain the direction.

In this case you've lost the sign of the north-south displacement. When the plane traveled southwards from B, what sign should you assign the displacement?

Your answer to (a) is good. Fix up the answer to (b) by setting the correct sign on the displacements. You've got the right ideas for the rest.
Wait, is (a) correct, then why did you say that I should change the sign of the displacement?

oh, ok, so the direction is actually -63.098 degrees because tan ^-1 (-814.2/413).

AidenPhysica said:
Wait, is (a) correct, then why did you say that I should change the sign of the displacement?
The signs of the components "disappear" when you square them using Pythagoras. So the sign didn't affect the result.

FYI, I changed the title of this thread to make it descriptive of the actual problem. By the forum rules, too vague or too general titles can be grounds for thread deletion, so please take the time to formulate thread titles that are descriptive of the question being asked.

## 1. What is displacement?

Displacement is a vector quantity that refers to the overall change in position of an object. It is the shortest distance between the starting and ending points of an object's motion.

## 2. How is displacement different from distance?

Distance is a scalar quantity that refers to the total path length covered by an object in motion. Displacement takes into account the direction of the motion and only considers the starting and ending points.

## 3. What is average velocity?

Average velocity is a vector quantity that refers to the overall rate of change in an object's position. It is calculated by dividing the displacement by the total time taken.

## 4. How is average velocity different from average speed?

While average velocity takes into account the direction of motion, average speed is a scalar quantity that only considers the magnitude of an object's motion. It is calculated by dividing the total distance traveled by the total time taken.

## 5. Can an object have a zero displacement but a non-zero average velocity?

Yes, an object can have a zero displacement but a non-zero average velocity if it changes direction during its motion. This means that the starting and ending points are the same, but the object's velocity is still changing over time.

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