KCET 2026 Physics Solved Paper

Concept:When velocity is given as a function of position $x$, acceleration is found using the chain rule: $a = v \frac{dv}{dx}$.Explanation:The velocity of the particle is:$$v = x^2 - 5x + 4$$To find the positions where velocity is zero, set $v = 0$:$$x^2 - 5x + 4 = 0$$Factorising:$$(x-1)(x-4) = 0$$So the positions are:$$x = 1 \ \mathrm{m} \quad \text{or} \quad x = 4 \ \mathrm{m}$$Now, acceleration is the rate of change of velocity with time:$$a = \frac{dv}{dt}$$Using the chain rule, since $v$ depends on $x$ and $x$ depends on $t$:$$a = \frac{dv}{dx} \cdot \frac{dx}{dt}$$But $\frac{dx}{dt} = v$, therefore:$$a = v \frac{dv}{dx}$$At the points where $v = 0$, the product $v \frac{dv}{dx}$ becomes:$$a = 0 \cdot \frac{dv}{dx} = 0$$Hence, the magnitude of acceleration when velocity is zero is:$$0 \ \mathrm{m/s^2}$$Answer:$$0 \ \mathrm{m/s^2}$$Correct option: C. Zero.