inaccensa

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At a certain time, race car has only a northward accelearation with a magnitutde of a1 and at a later time delta-t, the same race car has only an acceleration a2, what is the magnitutde of its average "j" at this time?

j= delta a/delta t

I solved the q and got the correct answer, but when I looked at princeton's vector diagram, I was a little thrown off.
Its a simple q asking to find the resultant change in acceleration

I drew an arrow pointing north (+a1) and then another from the head of the arrow pointing north eastward (+a2) to find the resultant a

Now, the explanation has an arrow eastward (+a2) and another from the head of the eastward arrow pointing south (-a1) I understand that there is no real difference between my and their vector diagram, but I'm a little weak at vectors, so want to be sure that my diagram was correct.

Secondly, they said a2-a1(incorrect answer choice) will be correct if accelerations given were vector accelerations, it is not true for magnitude of the accelerations???? :confused:
 

circulus vitios

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Scan a drawing or make something in mspaint. I don't understand what you're trying to say? :oops:
 
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circulus vitios

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Is that the entire question? If it is then I don't know why they drew the resultant vector like they did. But you are right, the magnitude of the resultant is the same. I'm not really sure what the second question means ("if accelerations given were vector accelerations"?) so I'll leave that for someone else.
 

geeyouknit

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Jun 22, 2011
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Secondly, they said a2-a1(incorrect answer choice) will be correct if accelerations given were vector accelerations, it is not true for magnitude of the accelerations???? :confused:
Not exactly sure what you mean? Can you clarify ? I think that they mean that you can't just subtract A2- A1 (like, 13 m/s2 - 5 m/s2) because these are vectors. Just subtracting would give you 8 m/s2. But you don't know the direction change.
Maybe they mean that it would be correct only if the vectors were pointed in the same direction? If A1 is north at 5 m/s2 and A2 is north at 13 m/s2, then you could say that the resultant is 8m/s2 / delta T.
 

inaccensa

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Not exactly sure what you mean? Can you clarify ? I think that they mean that you can't just subtract A2- A1 (like, 13 m/s2 - 5 m/s2) because these are vectors. Just subtracting would give you 8 m/s2. But you don't know the direction change.
Maybe they mean that it would be correct only if the vectors were pointed in the same direction? If A1 is north at 5 m/s2 and A2 is north at 13 m/s2, then you could say that the resultant is 8m/s2 / delta T.

hmmm true, but then again, the q had one vector pointing north and one east.

Im confused by the first part of your explanation, even though we know the direction and the magnitutde, we can't just subtract vectors, unless they are in the same direction...ie both north or both east, if one is north and the other is east, you'd have to use the pythagorean thm,correct?
 

geeyouknit

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Jun 22, 2011
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hmmm true, but then again, the q had one vector pointing north and one east.

Im confused by the first part of your explanation, even though we know the direction and the magnitutde, we can't just subtract vectors, unless they are in the same direction...ie both north or both east, if one is north and the other is east, you'd have to use the pythagorean thm,correct?
Yea. If you had A1 pointing north, and A2 pointing 30 deg east of north, you can't just subtract the magnitudes. If they are 90 degrees to each other, you can use pyth theroem. If they aren't, you can break them down into components of an xy axis.
 
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