Question

If the distance travel by a uniformly accelerated particle in $pth,qt$ and $rth$ second are $a,b$ and $c$ respectively. Then

• A. $(q-r)a+(r-p)b+(p-q)c=1$
• B. $(q-r)a+(r-p)b+(p-q)c=-1$
• C. $(q-r)a+(r-p)b+(p-q)c=0$
• D. $(q+r)a+(r+p)b+(p+q)c=0$

Answer 1

The correct option is C $(q-r)a+(r-p)b+(p-q)c=0$

Let $u$ be the initial velocity of the particle and $f$ be acceleration of the particle. Then
$a=u+\frac{1}{2}f(2p-1).....\left(i\right)$
$b=u+\frac{1}{2}f(2q-1)....\left(ii\right)$
$c=u+\frac{1}{2}f(2r-1).....\left(iii\right)$
Now, multiplying Eq. (i) by $(q-r)$, Eq. (ii) by $(r-p)$ and Eq. (iii) by $(p-q)$ and adding, we get $a(q-r)+b(r-p)+c(p-q)=a\left(0\right)+\frac{1}{2}f\left(0\right)$
$\therefore a(q-r)+b(r-p)+c(p-q)=0$.