Porcia C.

asked • 10/16/16

help!!! i dont get the last fewparts of the question

A fish swimming in a horizontal plane has velocity vi = (4.00 i + 3.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri = (10.0 i + 4.00 j) m. After the fish swims with constant acceleration for 23.0 s, its velocity is v = (17.0 i - 5.00 j) m/s.
(a) What are the components of the acceleration?
ax =
.57

Correct: Your answer is correct.
m/s2
ay =
-.35

Correct: Your answer is correct.
m/s2

(b) What is the direction of the acceleration with respect to unit vector i?

-31.55

Correct: Your answer is correct.
° (counterclockwise from the +x-axis is positive)

(c) If the fish maintains constant acceleration, where is it at t = 29.0 s?
x =
365.69

Correct: Your answer is correct.
m
y =
-64.175

Incorrect: Your answer is incorrect.

Your response differs from the correct answer by more than 10%. Double check your calculations. m
In what direction is it moving?

-

Incorrect: Your answer is incorrect.
° (counterclockwise from the

1 Expert Answer

By:

Porcia C.

What about the degrees counter-clockwise from the x axis
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10/16/16

Steven W.

tutor
It looked like you computed the angle correctly in (a), so -- once you have the correct components for both x and y -- identify which is opposite and which is adjacent with respect to an angle measured from the +x axis to the resultant.  Then compute the angle using the standard opposite/adjacent formula, just as you appear to have done correctly before.  See if that works!
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10/16/16

Porcia C.

I tried. Physics just puzzles me lol I'm so. Lost
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10/16/16

Steven W.

tutor
Well, it looks like you are getting a lot of it correct.  Your only error above was mathematical.
 
So, the x component of its position is (+)365.7, as you correctly determined
The y component, which you almost correctly determined, is -56.175.
 
So the position must be in the 4th quadrant.  If you look at a right triangle with those components as the legs, the hypotenuse points into the 4th quadrant, and the angle it makes with the +x axis is such that the y component is opposite the angle, and the x component is adjacent.
 
So, you can compute the smallest angle between the +x axis and the position vector by using
 
tan(θ) = (56.175)/(365.69) --> θ = tan-1(56.175/365.69) = 8.73°
 
And this is BELOW the +x axis, or, in other words, measured clockwise from it.  To get the corresponding counterclockwise angle, we would have to imagine starting at the +x axis, and swing all the way around into the 4th quadrant to match up with the vector.  You may be able to see that this counterclockwise angle is just 360° - 8.73° = 351.27°.
 
You can think of this as, we swing all the way around counterclockwise, and then back up clockwise by the 8.73° we calculated above to line with the position vector.
 
Keep the questions coming if you are still lost in any way!
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10/16/16

Porcia C.

It says that answer wrong
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10/16/16

Steven W.

tutor
Hmmm, that is from the +x axis counterclockwise.  It said your answer of -31.55 for the acceleration direction above was correct.  So maybe try -8.73.
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10/16/16

Porcia C.

still wrong 
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10/16/16

Steven W.

tutor
Then, for the moment, I am a bit flummoxed, because (except for that math error), I agree with your numbers. I will look at it again when I get home. 
 
Could it possibly be a significant figures issue?
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10/16/16

Steven W.

tutor
No, I still get the y-direction position is -56.175, and then the same value for the angle:  8.73 degrees below the +x axis. and thus 351.27 measured counterclockwise.
 
The only thing I can think of is that it needs fewer significant figures or something.  But, in terms of calculation, I cannot find anything wrong with it, especially given that your components of acceleration are right on.
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10/16/16

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