Question

A particle in uniformly accelerated motion travels $a,b$ and $c$ distances in the $x^{th},y^{th}$ and $z^{th}$ second of it's motion, respectively. Then, value of $a(y-z)+b(z-x)+c(x-y)$ will be

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (B)

0

Let the uniform acceleration be $p$ and for convenience, let the initial velocity be $u$ .
Then , using $s_n=u+\frac{1}{2}a(2t-1)$
We have , $a=u-\frac{p}{2}+px$
$b=u-\frac{p}{2}+py$
$c=u-\frac{p}{2}+pz$
$a(y-z)+b(z-x)+c(x-y)=pxy-pxz+pyz-pxy+pxz-pxy=0$