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A particle leaves the origin with an initial velocity v Overscript right-arrow EndScripts equals left-parenthesis 8.47...

A particle leaves the origin with an initial velocity v Overscript right-arrow EndScripts equals left-parenthesis 8.47 i Overscript ̂ EndScripts right-parenthesis m divided by s and a constant acceleration a Overscript right-arrow EndScripts equals left-parenthesis negative 1.24 i Overscript ̂ EndScripts minus 4.99 j Overscript ̂ EndScripts right-parenthesis m divided by s Superscript 2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector?

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Part-(a) The acceleration is given as ä- (-1.241 -4.99j)m/s2 The horizontal component of the acceleration is, q = (-1241 )m/s2 The horizontal component of the initial velocity is, Vox-(8.47i)m/s When the particle reaches maximum x-coordinate ½-0 The kinematic equation used to find the final velocity after some time t is 0x 0-(847m/s)+(-124m/s) 0 (8.47m/s)+(-1.24 m/s2 )r 8.47m/s 1.24 m/s2 t 6.83s The time taken to reach the maximum x coordinate is t 6.83s The kinematic equation used to find the final velocity after some time t is, v-(8.47î m/s+(-1.24i -4.99j)m/s (6.83 s v =|((0.0008)^-(34.1) j +0k) m/s

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