140_Physics ProblemsTechnical Physics

140_Physics ProblemsTechnical Physics - 142 P5.52 The Laws...

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142 The Laws of Motion P5.52 F a = m gives the object’s acceleration a ij ai j v == =− = F m t t d dt 800 400 200 . ± . ± . ± . ± . ej N 2.00 kg ms 23 Its velocity is dd t td t tt v v i t t i vvv v a vi j j zz z =− =−= 0 100 0 0 2 . ± . ± . ± . ± . ±ms e j (a) We require v = 15 0 . ms , v 2 225 = 22 16 0 1 00 225 160 225 0 16 0 16 0 4 225 900 300 24 42 2 2 .. . . t t s s s s 26 2 += +− = = −± = = a f a f Take r i = 0 at t = 0. The position is rv i j rij dt t t dt 0 2 0 2 3 . ± . ± . ± . ± e j at t = 3 s we evaluate. (c) ri j 18 0 9 00
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