An object with an initial velocity undergoes constant acceleration.Position info

Question

An object with an initial velocity undergoes constant acceleration.Position information was collected at four points and is shown onthe table. Using yn+1-yn =vnt+1/2ayt2 and yn+2-yn=vnt+1/2ayt
determine the acceleration and fill each box. time(s) position (m)    average accl (m/s2 1    8    ___________ 2    25    ___________ and yn+2-yn=vnt+1/2ayt
determine the acceleration and fill each box. time(s) position (m)    average accl (m/s2 1    8    ___________ 2    25    ___________ determine the acceleration and fill each box. time(s) position (m)    average accl (m/s2 1    8    ___________ 2    25    ___________

Explanation / Answer

The first equation is yn + 1 - yn = vn * t +(1/2)ay * t2 and the second equation is yn + 2 - yn = vn * t +(1/2)ay * t2 and the second equation is yn + 2 - yn = vn * t +(1/2)ay * t2 yn + 1 - yn = 8 m when t = 1 s or 8 = vn * t + (1/2)ay * t2= vn * 1 + (1/2)ay * (1)2 or vn + (ay/2) = 8 ----------(1) yn + 2 - yn = 25 m when t = 2s or 25 = vn * t + (1/2)ay * t2= vn * 2 + (1/2)ay *(2)2 or 2vn + 2ay = 8 or vn + ay = 4 ----------(2) solving equaions (1) and (2) we get ay/2 = -4 or ay = -8 m/s2 and vn = 8 - (ay/2) = 8 - (-4) = 12m/s and vn = 8 - (ay/2) = 8 - (-4) = 12m/s

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