Question

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 $(q-r)x+(r-p)y+(p-q)z=0$

Answer 1

As $S_{nth}=u+an-\frac{1}{2}a=u+\frac{a}{2}(2n-1)$
$x=u+\frac{a}{2}(2p-1)$... (i)

$y=u+\frac{a}{2}(2q-1)$... (ii)
$z=u+\frac{a}{2}(2r-1)$... (iii)
Subtracting Eq. (iii) from Eq. (ii),
$y-z=\frac{a}{2}(2q-r)$ or $q-r=\frac{y-z}{a}$
or $(q-r)x=\frac{1}{a}(yx-zx)$ ..(iv)
Similrly, we can show that
$(r-p)y=\frac{1}{a}(yx-zx)$ ..(v)
and $(p-q)z=\frac{1}{a}(xz-yz)$ ..(vi)
Adding Eqs. (iv), (v) and (vi), we get
$(q-r)x+(r-p)y+(p-q)z=0$