Vector description of motion around a bend

  • Thread starter ianperez
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  • #1
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Homework Statement


A car, initially going eastward, rounds a 90 degree bend and ends up heading southward. If the speedometer reading remains constant, what is the direction of a car's average acceleration vector?

-So i know the speed remains constant.

Homework Equations



average acceleration =( final velocity - initial velocity ) / elapsed time

The Attempt at a Solution



initial velocity = vi + 0j
final velocity = 0i + -vj
average acceleration = (-vi + -vj) / elapsed time

I don't think that the time matters here, because the ratio of vj/vi will remain the same, so:

direction of motion = inverse tangent of (-vj/-vi) = 45 degrees.

I feel like the answer should be -45 degrees. Also the back of the book says average acceleration = 5.12 m/s2 and that the angle should be 41 degrees south of west. How can they get an actual value for average acceleration? Any help would be greatly appreciated.
 

Answers and Replies

  • #2
OldEngr63
Gold Member
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Find the instantaneous acceleration as a V^2/R calculation, with appropriate direction. Then integrate over the whole process and divide by the length of the interval to get a vector average.
 
  • #3
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4,725

Homework Statement


A car, initially going eastward, rounds a 90 degree bend and ends up heading southward. If the speedometer reading remains constant, what is the direction of a car's average acceleration vector?

-So i know the speed remains constant.

Homework Equations



average acceleration =( final velocity - initial velocity ) / elapsed time

The Attempt at a Solution



initial velocity = vi + 0j
final velocity = 0i + -vj
average acceleration = (-vi + -vj) / elapsed time

I don't think that the time matters here, because the ratio of vj/vi will remain the same, so:

direction of motion = inverse tangent of (-vj/-vi) = 45 degrees.

I feel like the answer should be -45 degrees. Also the back of the book says average acceleration = 5.12 m/s2 and that the angle should be 41 degrees south of west. How can they get an actual value for average acceleration? Any help would be greatly appreciated.
You had the equation for the average acceleration correct. Now, what do you get when you add -vi and -vj? Draw a diagram and see.

That 41 has to be a typo. It has to be 45.

Chet
 
  • #4
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Find the instantaneous acceleration as a V^2/R calculation, with appropriate direction. Then integrate over the whole process and divide by the length of the interval to get a vector average.

That sounds like a lot of fun, but I'm certain that's beyond my present day abilities. Thanks though
 
  • #5
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You had the equation for the average acceleration correct. Now, what do you get when you add -vi and -vj? Draw a diagram and see.

That 41 has to be a typo. It has to be 45.

Chet

Thanks for your help. When I draw it out I see an angle that is 45 degrees south of west, which makes sense to me. I'm really glad to hear you say the 41 degrees is a typo, but what about the magnitude of average acceleration of 5.12m/s sq. That has to be a typo too right? Since the question only asked for the direction?
 
  • #6
BvU
Science Advisor
Homework Helper
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The question only asks for the direction.

If you draw a diagram, you see that the vector addition of vf and -vi gives a vector of length v √(½) "direction" SW.
To convert that into an average acceleration, you need to assume something, e.g. a constant acceleration during the turn. That gives a quarter circle trajectory of length ½π r/v, so the average acceleration is v √(½) / (½π r/v) = v2√(2)/(πr). I don't see a way to extract a number from that without further information.
 
  • #7
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4,725
It looks like the problem statement left out some info. Why don't you take their number for accel magnitude, and back out what the velocity they meant to include should have been?

Chet
 
  • #8
nasu
Gold Member
3,777
433
Maybe you are looking at the answer for a different problem. This one does not ask about magnitude of acceleration. Unless is not the complete text.
 

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