Q. In some appropriate units, time ($t$) and position ($x$) relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is
A) $-\frac{2}{(x+2)^3}$
B) $-\frac{2}{(2x+1)^3}$
C) $+\frac{2}{(x+1)^3}$
D) $+\frac{2}{2x+1}$
Explanation
[Solution by Claude - not from source] $\frac{dt}{dx}=2x+1\Rightarrow v=\frac{1}{2x+1}$. $a=v\frac{dv}{dx}=\frac{1}{2x+1}\cdot\frac{-2}{(2x+1)^2}=-\frac{2}{(2x+1)^3}$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.