20&amp;23 - 29 29 29 29 2 4 cos 20 4 sin 20 s m t t dt...

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TAM 212 Spring, 2004 1.20 Given: ( 29 ( 29 ( 29 j i v 4 / cos 20 4 / sin 20 / t t F P p p - = m/s, where i and j , are unit vectors fixed in the frame of reference. t is time measured in seconds. Required: ( 29 ) 2 / s t F P = a Solution: Again, since we are only worried about one frame of reference, we suppress it in this problem. First, again, we solve as a function of t, and then plug in the specifics.
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Unformatted text preview: ( 29 ( 29 ( 29 ( 29 2 / / 4 / cos 20 4 / sin 20 / s m t t dt d t F P j i a p p-= (Remember, the extra factor of 1/s comes from the derivative and the definition of t as in s.) So… ( 29 ( 29 ( 29 ( 29 2 / / 4 / sin 5 4 / cos 5 s m t t t F P j i a p p p p + = 1.23 Done in class ( 29 2 / / 5 2 s m s t F P j a p = =...
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