Q. A particle moving in a straight line with uniform acceleration is observed to be a distance $a$ from a fixed point initially. It is at distances $b,c,d\ $from the same point after $n,2n,3n$ second. The acceleration of the particle is
A) $\frac{c - 2b + a}{n^{2}}$
B) $\frac{c + b + a}{{9n}^{2}}$
C) $\frac{c + 2b + a}{{4n}^{2}}$
D) $\frac{c - b + a}{n^{2}}$
Explanation
(a) $b - a = un + \frac{1}{2}An^{2}$
$2b - 2a = 2un + An^{2}$
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Again, $c - a = u(2n) + \frac{1}{2}A(2n)^{2}$
Subtracting, $c - a - 2b + 2a = An^{2}$
$A = \frac{c - 2b + a}{n^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.