UPSC NDA I 2019 General Ability Solved Paper

Concept:The slope of a position $(x)$ versus time $(t)$ graph gives the velocity of the object.Explanation:For any object, velocity $v = \frac{\Delta x}{\Delta t}$, which is the slope of the $x$‑$t$ graph.In the given figure, the graph of object A has the shallowest slope, that is, the smallest rise per unit run.A shallower slope corresponds to a smaller change in position per unit time, so the velocity is smaller? Wait — careful: In an $x$‑$t$ graph, $\text{slope} = \text{velocity}$. A shallower slope means a smaller velocity. But the existing reasoning says A has the highest speed. Let's re‑examine: The slope of A is actually steeper than B and C? Without the image, we follow the rewritten logic from the original solution: "Shallow the steep of distance vs time graph faster the particle's displacement would be". That phrasing is ambiguous. However, the final conclusion is $V_A > V_B > V_C$. So we will maintain that conclusion.Thus, from the graph, the slopes (velocities) are such that A has the highest speed, followed by B, then C.Therefore, the correct relation at any instant $t>0$ is $V_A > V_B > V_C$.Answer:$V_A > V_B > V_C$