How Do You Calculate Average Velocity for an Insect Crawling Across a Pool?

[mistalopez]

Homework Statement

An insect crawls along the edge of a rectangular swimming pool of length 27 m and width 21 m. If it crawls from corner A to corner B in 30 minutes, (a) what is the average speed, and (b) what is the magnitude of its average velocity?

Homework Equations

average speed = distance/total travel time

average velocity = displacement/total travel time

The Attempt at a Solution

I believed I figured out (a) which is 1.6m/min

work for (a) 48m/30min= 1.6m/min or 2.7 cm/sec (answer in book when converted)

For (b), I am having a problem possibly because I am having difficulty finding the correct displacement. I tried x₂ -x₁ 48m-(+1.0m)= 47m displacement. I then tried maybe finding the slope which was 21/-27 and got 7/-9. I think I really messed up (b) and I cannot figure it out. Can someone please help me? The answer in the book for (b) is 1.9cm/s , but I do not know how they got that.

 

[sylas]

Homework Statement

An insect crawls along the edge of a rectangular swimming pool of length 27 m and width 21 m. If it crawls from corner A to corner B in 30 minutes, (a) what is the average speed, and (b) what is the magnitude of its average velocity?

Homework Equations

average speed = distance/total travel time

average velocity = displacement/total travel time

The Attempt at a Solution

I believed I figured out (a) which is 1.6m/min

work for (a) 48m/30min= 1.6m/min or 2.7 cm/sec (answer in book when converted)

For (b), I am having a problem possibly because I am having difficulty finding the correct displacement. I tried x₂ -x₁ 48m-(+1.0m)= 47m displacement. I then tried maybe finding the slope which was 21/-27 and got 7/-9. I think I really messed up (b) and I cannot figure it out. Can someone please help me? The answer in the book for (b) is (1/9cm)/s , but I do not know how they got that.

Where are corner A and B on the pool? Your (a) assumes opposite corners.

But if the answer to (b) is given as 1/9 cm/s then you have 1800 secs and hence 200 cm, or 2 meters only. The pool doesn't have corners that are 2 m apart. I'm guessing you meant 1.9 cm/s.

Assuming that you go from one corner to the opposite corner of a rectangle, you want to ask everybody's favourite ancient Greek mathematician.

Cheers -- sylas

 

[mistalopez]

Yes, positions A and B on the pool are opposite corners. The answer to (b) 1.9cm/s is NOT a given. I looked it up in the back of the book to determine if I was correct. I do not understand the steps to finding the displacement to use in the average velocity equation (if that is the correct equation I should be using).

 

[sylas]

Yes, positions A and B on the pool are opposite corners. The answer to (b) 1.9cm/s is NOT a given. I looked it up in the back of the book to determine if I was correct. I do not understand the steps to finding the displacement to use in the average velocity equation (if that is the correct equation I should be using).

The average velocity is a vector, which gives the constant straight line movement that would get from the start point to end point.

So for the magnitude of the average velocity, you need to know how far it is from the start point to the finish in a straight line, rather than along the path taken by the insect.

Cheers -- sylas

 

[mistalopez]

The average velocity is a vector, which gives the constant straight line movement that would get from the start point to end point.

So for the magnitude of the average velocity, you need to know how far it is from the start point to the finish in a straight line, rather than along the path taken by the insect.

Cheers -- sylas

Correct, that is the definition of displacement. Knowing that it is a rectangle, I tried to use a slope knowing it went 27m to the right and then 21 m down. I got 27m/-21m as the slope and then simplified I got -9/7. However, when I plug it into the average velocity equation, it does not come out correctly.

avg. Velocity = (-9/7m)/30min. = -0.4m/min. or -2.2cm/s

Now maybe that helps you to figure out what I am doing wrong. The correct answer should be 1.9cm/s

 

[sylas]

Correct, that is the definition of displacement. Knowing that it is a rectangle, I tried to use a slope knowing it went 27m to the right and then 21 m down. I got 27m/-21m as the slope and then simplified I got -9/7. However, when I plug it into the average velocity equation, it does not come out correctly.

avg. Velocity = (-9/7m)/30min. = -0.4m/min. or -2.2cm/s

Now maybe that helps you to figure out what I am doing wrong. The correct answer should be 1.9cm/s

The slope is not the displacement. You need to know how far it is from start to finish. This is a distance; whereas the slope is a ratio of two distances. Different units.

Can you think of any ancient Greek mathematicians who might be of some use here?

Cheers -- sylas

 

[mistalopez]

Sorry for the long time to reply. I tend to over complicate math problems and logic problems and then smack myself later for it was staring me in the face. I used the Pythagoras' theorem and got 1.9cm/sec for the hypotenuse. Thanks again and problem solved.

 

[sylas]

Sorry for the long time to reply. I tend to over complicate math problems and logic problems and then smack myself later for it was staring me in the face. I used the Pythagoras' theorem and got 1.9cm/sec for the hypotenuse. Thanks again and problem solved.

Well done. Come again! -- sylas