Tangential and Radial Acceleration of car around a curve

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Homework Help Overview

The discussion centers around a problem involving the tangential and radial acceleration of a car navigating a highway curve, specifically analyzing the average acceleration as the car's speed doubles between two points.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of average acceleration and attempt to determine the direction of the average acceleration vector based on changes in velocity. There are considerations of vector diagrams and angles related to the resultant vector.

Discussion Status

Participants are actively engaging with the problem, drawing conclusions about the direction of the average acceleration and questioning the validity of their options. There is a back-and-forth regarding the angles associated with the options presented, indicating a productive exploration of the topic.

Contextual Notes

Some participants express uncertainty about the angles and directions involved, and there is a mention of avoiding Cartesian coordinates in favor of cardinal directions. The discussion reflects a collaborative effort to clarify the problem setup and assumptions.

123scope
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Hi, I have a problem with this problem..
The figure (in the attachment) shows a bird's-eye view of a car going around a highway curve. As the car moves from point 1 to point 2, its speed doubles. Which vector shows the direction of the car's average acceleration between these two points?

Ok. According to my knowledge,
THe answer cannot be (g) and (b) and (a) is also unlikely
Applying the radial and tangential acceleration I guess it could be (d) and (f)
But I can't make my mind... Can anyine help me.
 

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Start with the definition of average acceleration: \Delta{\vec{v}}/\Delta t.

Draw the initial velocity vector. Draw the final velocity vector. The change in velocity is the difference between those two vectors. That should enable you to choose the correct answer.

AM
 
So using a = (vf - vi) / (delta t)

I get a = (-2v i - v j) / (delta t)

So does that mean the direction of the average acceleration is in the third quadrant??
 
123scope said:
So using a = (vf - vi) / (delta t)

I get a = (-2v i - v j) / (delta t)

So does that mean the direction of the average acceleration is in the third quadrant??
Don't bother putting them on an cartesian plane - just use North/South etc. The initial velocity vector is 1 unit North and the final velocity is 2 units west. Where does the difference vector point?

AM
 
The vector diagram is as follows/
I found it to be south west. So could it be option (g)?
 

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123scope said:
The vector diagram is as follows/
I found it to be south west. So could it be option (g)?
What is the angle (south of west) that the resultant points in your diagram? What is the angle in g)? (You can measure the latter if you can't tell just by looking at it).

AM
 
sorry woops. it the answer can't be option (g). THe angle in option (g) is slightly larger.
The answer should be (f) then.
26.57 degrees south of west. Am I right?
 
123scope said:
sorry woops. it the answer can't be option (g). THe angle in option (g) is slightly larger.
The answer should be (f) then.
26.57 degrees south of west. Am I right?
Right.

AM
 

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