Question regarding position/displacement vectors

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Homework Help Overview

The problem involves calculating the displacement of an object (a bee) over a short time interval using position vectors at specific times. The context is rooted in kinematics, specifically focusing on displacement and average velocity calculations.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the average velocity calculated from earlier parts of the question to find the displacement over a very short time interval. Some participants question the validity of assuming the average velocity remains constant over this interval. Others suggest using the average velocity directly for the time interval to calculate displacement.

Discussion Status

Participants are exploring different methods to calculate displacement, with some finding alternative results based on their approaches. There is a recognition of differing answers, and some participants express gratitude for the insights shared by others, indicating a collaborative effort to clarify the problem.

Contextual Notes

There is an implicit assumption about the constancy of average velocity over the short time interval, which is being questioned. The discussion also reflects on the appropriateness of the calculated magnitudes of displacement in relation to the overall motion described.

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Homework Statement



The position of an object at three different times is given

At t= 4.9s
position= (4.0,3.1,0) m

At t= 5.4s
position = (1.3, 7.1, 0) m

At t= 5.9s
position = (-0.8,12.7,0) m

Using the best information available, what was the displacement of the bee during the time interval from 4.9 s to 4.93 s?

The earlier parts of this question asked for the average velocity of the object between 4.9s and 5.4 s (-5.4,8,0) m/s

And the average velocity between 4.9s and 5.9 s which is (-4.8, 9.6, 0) m/s.

Homework Equations



Displacement formula = r2 - r1 where r2 is the final position of the object and r1 is the initial position

and also,

r2 = r1 + (Average Velocity)(change in time)

The Attempt at a Solution



The way I attemped this solution is to use the average velocity between 4.9 s and 5.4 s (which was found earlier) and plug all the given values into the formula:

r2= r1 + (Average Velocity)(Change in time)

Which would be, (4.0,3.1,0) + ((-5.4,8,0)(0.93))
= (-1.022, 11.1, 0) = r2

I would then use this value for r2 into the formula for finding the displacement (r2 - r1)

(-1.022,11.1,0) - (4.0,3.1,0)

= (-5.022, 8, 0) m

So the final answer would be (-5.022, 8, 0) m

However, I'm still kind of sketchy on this solution and would just like some confirmation whether this is the right way of doing a question like this.
 
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I get a different answer ..

Assuming the average velocity is the same for the interval 4.9 to 4.93 s
as it is for the interval 4.9 to 5.4 s ..

displacement for the interval 4.9 to 4.93 s =
(average velocity)(time interval) =
(-5.4,8,0) m/s multiplied by (4.93 s - 4.9 s)
 
mikelepore said:
I get a different answer ..

Assuming the average velocity is the same for the interval 4.9 to 4.93 s
as it is for the interval 4.9 to 5.4 s ..

displacement for the interval 4.9 to 4.93 s =
(average velocity)(time interval) =
(-5.4,8,0) m/s multiplied by (4.93 s - 4.9 s)


Wow mikelepore, I totally forgot about that equation for some reason. Thanks, the answer I got from your method is

(-0.162,0.24,0)

which makes much more sense. I just realized the magnitude for the displacement vector I had initially calculated was about 9.44 m which is almost double the magnitude of the displacement for the time interval between 4.9s and 5.4s, which doesn't seem to make sense for a question like this.

Thanks again.
 
Now I got the same answer as you.
 

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