CHAPTER 3 - 9

# CHAPTER 3 - 9 - 24 Theaccelerationisa=(3.0m/s)i(4.5m/s)j 2...

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d v v 0 v = a d t 0 t ; v – (5.0 m/ s) i = (– 3.0 m/ s 2 ) i + (4.5 m/ s 2 ) j d t 0 t = (– 3.0 m/ s 2 ) t i + (4.5 m/ s 2 ) t j , or       d r 0 r = v d t 0 t ;       r = (– 3.0 m/ s 2 ) t + 5.0 m/ s i + (4.5 m/ s 2 ) t d t 0 t = 1 2 (– 3.0 m/ s 2 ) t 2 + (5.0 m/ s) t i + 1 2 (4.5 m/ s 2 ) t 2 j = (– 1.5 m/ s 2 ) t 2 + (5.0 m/ s) t i + (2.25 m/ s 2 ) t 2 j . 24. The acceleration is  a  = (– 3.0 m/s 2 ) i  + (4.5 m/s 2 ) j . We find the velocity by integrating: v  = [(– 3.0 m/s 2 ) t  + 5.0 m/s] i  + (4.5 m/s 2 ) t j . We find the position by integrating: To find the time at which the particle reaches its maximum  x -coordinate, we set d x /d t  = 0: d x /d t  = (– 3.0 m/s 2 ) t  + 5.0 m/s = 0, which gives  t  = 1.67 s. The velocity is v  = [(– 3.0 m/s 2 ) t  + 5.0 m/s] i  + (4.5 m/s 2 ) t j  = 0 i  + (4.5 m/s 2 )(1.67 s) j  =        (7.5 m/s) j . The position is

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CHAPTER 3 - 9 - 24 Theaccelerationisa=(3.0m/s)i(4.5m/s)j 2...

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