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A particle is moving in a straight line with constant acceleration. If x,y and z be the distances described by a particle during the pth, qth and rth second respectively, prove that (q-r)x+(r-p)y+(p-q)z=0

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To solve the problem, we need to prove that: \[ (q - r)x + (r - p)y + (p - q)z = 0 \] where \(x\), \(y\), and \(z\) are the distances covered by a particle during the \(p\)th, \(q\)th, and \(r\)th seconds respectively, under constant acceleration. ...
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