Dynamics: Relative Acceleration

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Master1022
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Homework Statement


The problem is in the image attached below
Textbook diagram for Problem 2/241 showing a street grid with north arrow, car B heading south 30 m from an intersection, and car A on a 60 m radius curve; a 30 degree angle is marked between the lines to each car.


2. Homework Equations


In the working below (although I think I have misunderstood a concept)

The Attempt at a Solution


My attempt is in the attached image.
Handwritten working for a relative acceleration problem: a_A/B = a_A - a_B, with polar acceleration components and a vector triangle showing -a_B, a_A/B and a 30 degree angle.


I tried to leave the numbers out until the end. I have not written the exact working as I did it in draft elsewhere. I used the Law of Cosines on the vector triangle, in order to obtain [itex]\vec v_{A/B}[/itex], which came out as 2.057544... m/s. But that is not the correct answer.

Upon reflection, I have realized that I was supposed to draw the vector triangle such that the angle between [itex]- \vec v_{B}[/itex] and [itex]\vec v_{A}[/itex] is obtuse. How should one decide on how to draw the vector triangle?

Many thanks.
 
Attachments
  • Handwritten vector triangle sketch with labelled vectors v_A and v_B drawn from a common origin, captioned Velocity Diagram; small image on a printed page.
    image.png
    2.3 KB · Views: 351
  • Handwritten working for a relative acceleration problem: a_A/B = a_A - a_B, with polar acceleration components and a vector triangle showing -a_B, a_A/B and a 30 degree angle.
    IMG_6186.jpg
    43.7 KB · Views: 568
  • Textbook diagram for Problem 2/241 showing a street grid with north arrow, car B heading south 30 m from an intersection, and car A on a 60 m radius curve; a 30 degree angle is marked between the lines to each car.
    IMG_6185.JPG
    47.7 KB · Views: 729
Last edited:
You need to construct ##\vec a_A - \vec a_B##, which is the same as ##\vec a_A + (- \vec a_B)##. So, your diagram should show ##\vec a_A## and ## (- \vec a_B)## added together.
 
TSny said:
You need to construct ##\vec a_A - \vec a_B##, which is the same as ##\vec a_A + (- \vec a_B)##. So, your diagram should show ##\vec a_A## and ## (- \vec a_B)## added together.
Thanks for your response, it makes sense now.