Selected Hw Problems for Ch. 3 and 4

# Selected Hw Problems for Ch. 3 and 4 - 3.7 z = 3.00 m a...

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3.7 x z y z = 3.00 m x = 3.70 m y = 4.00 m fly (start) fly (end) a) What’s the magnitude of the fly’s displacement? This is the length of the (red) displacement vector from the origin (where the fly started) to the opposite corner (where the fly ended). i ˆˆ ˆ r0 i 0 j 0 k = ++ r Putting the fly’s initial position at the origin of our coordinate system makes its initial position vector, This sensible choice of the coordinate origin makes the vector from the origin to the diagonal corner the displacement vector of interest. In this coordinate system this vector is, fi ˆˆˆ d r r (3.70 m)i (4.30 m)j (3.00 m)k =−= + + r rr (which answers part f) It’s length is, 222 2 2 2 fx y z d r r r r (3.70m) (4.30m) (3.00m) 6.42m == = + + =

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To be clear about what we mean by sensible above let’s consider another choice for our coordinate origin. x y fly (start) fly (end) z Now, i ˆˆ ˆ r 0i ( 4.30 m)j 0k =+ + r f ˆ r (3.70 m)i 0j (3.00 m)k + r And, fi ˆ ˆ ˆ ˆ d r r (3.70 m)i 0i =−= + + + + r rr ˆˆˆ d r r (3.70 m)i (4.30 m)j (3.00 m)k + + r The same as for our original choice of origin as it should. If we were true masochists we could have also rotated the coordinate system with respect to the room. The displacement vector would not be changed (though our leisure time would be). The moral is a little forethought can save you effort.
b) No, there is no shorter path between opposite corners than the diagonal (a straight line is the shortest path between two points). c) Yes. It’s path could be greater (it. could e.g. zigzag off and on the diagonal line). d) Yes. It’s path could be equal by taking the diagonal path f) answered in part a) g) This is tricky. Consider unfolding the room and having the fly walk along the straight line shortest path: 22 W d (3.7 m 3.0 m) (4.30 m) 7.96 m =+ + = x z y Bottom wall front wall end start 3.70 m 3.00 m 4.30 m In this case,

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3.22 He must now go the displacement vector from where he ended up to where he wanted to be fi dr r =− r rr Where f ˆ r (5.8 km)j = r We must express in unit vector notation (see figure), Impose a coordinate system so that the & directions lie along East and North respectively, Having gone the vector ˆ i ˆ j o i r (7.8 km) 50 N =∠ r i r r r θ rsin θ rcos θ oo i ˆˆ r (7.8 km) cos(50 )i (7.8 km) sin(50 )j =+ r i r (5.01km)i (5.98 km) j r i r r f r r d r
Then, fi ˆˆ ˆ ˆ ˆ d r r (5.6 km)j (5.01km)i (5.98 km) j ( 5.01km)i ( 0.38 km)j ⎡⎤ =−= + = + ⎣⎦ r rr The magnitude of this is i r r f r r d r 22 2 2 xy d d d (5.01km) (0.38 km) 5.0 km =+ = + = To get the angle, translate to the origin. Then with respect to the negative x axis consider the figure to the right in which the angle has been exaggerated.

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Selected Hw Problems for Ch. 3 and 4 - 3.7 z = 3.00 m a...

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