Velocity-dependent Acceleration

When acceleration depends on velocity rather than time, the constant-acceleration equations do not apply and one must integrate the differential relation directly. To connect velocity with distance it is efficient to write acceleration as $a=v\frac{dv}{dx}$, the velocity-position form. Here $v\frac{dv}{dx}=-2v$, which simplifies to $\frac{dv}{dx}=-2$ for $v\ne 0$. This shows velocity decreases linearly with position. Integrating from the initial velocity $10\text{ m/s}$ at $x=0$ down to $v=0$ gives ${\int }_{10}^{0}dv={\int }_0^x-2dx$, so $0-10=-2x$, yielding $x=5\text{ m}$. The option 10 m is wrong because it equates distance numerically with the initial speed. The option 2.5 m halves the correct result by mishandling the factor of two. The option 20 m doubles it by integrating incorrectly. This is the JEE Advanced calculus approach to velocity-dependent acceleration. As a check, although the particle takes infinite time to truly stop under exponential decay $v=10e^{-2t}$, the total distance converges to a finite 5 m, consistent with the integral of the decaying velocity, ${\int }_0^{\infty }10e^{-2t}dt=5\text{ m}$.