Velocity of a ladybug relative to the ground

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Homework Help Overview

The problem involves determining the velocity of a ladybug relative to the ground, given its velocity and the velocity of the chair it is on. The subject area pertains to vector addition and motion in two dimensions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the y-component and x-component of the chair's velocity, as well as the addition of these components to find the resultant velocity. There is also a focus on determining the direction of the resultant velocity.

Discussion Status

Some participants confirm the calculations presented, while others suggest clarifying the direction of the velocity. The discussion includes multiple interpretations of how to express the direction, indicating a productive exploration of the problem.

Contextual Notes

Participants note constraints such as significant figures and the preferred notation for direction in relation to the original problem statement.

TheronSimon
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Homework Statement


A ladybug with a velocity of 10 mm/s [W] crawls on a chair that is being pulled with a velocity of 40 mm/s [40 degrees N of W]. What is the velocity of the ladybug relative to the ground?


Homework Equations





The Attempt at a Solution


finding the y-component
40(sin40) = 25.7

x-component requires the addition of two x-axis component vectors
40cos40 + 10 = 40.6

Now a² + b² = c²
25.7² + 40.6² = c²
c = 48.1mm/s
 
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Hi TheronSimon! :smile:
TheronSimon said:
A ladybug with a velocity of 10 mm/s [W] crawls on a chair that is being pulled with a velocity of 40 mm/s [40 degrees N of W]. What is the velocity of the ladybug relative to the ground?

finding the y-component
40(sin40) = 25.7

x-component requires the addition of two x-axis component vectors
40cos40 + 10 = 40.6

Now a² + b² = c²
25.7² + 40.6² = c²
c = 48.1mm/s

looks ok :smile:

now you just need to state the direction of the velocity :wink:

(btw, maximum 2 significant figures in this case :wink:)
 
so for the direction tan -1 (25.7/40.6) = 32'

so the complete answer is 48 mm/s [W32'N] ?
 
yes :smile:

(probably best to call it "32° N of W", same as in the question :wink:)

(btw, you'll find a ° in the Quick Symbols box on the right of the Reply box)
 
alright thank you very much Tim! :)
 

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