UPSC CDS 1 2023 GK Paper

Concept:The area under the velocity-time graph for a particle moving in a straight line with uniform acceleration gives the net displacement.Explanation:For a particle undergoing uniform acceleration, its velocity changes at a constant rate.The velocity-time graph is a straight line.The area under the graph between two time points equals the integral $\int v \, dt$, which physically represents the change in position, i.e., displacement.In the case of uniform acceleration, this area is a trapezoid.The area is calculated as $\frac{1}{2}(u+v)t$, where $u$ is initial velocity, $v$ is final velocity, and $t$ is the time interval.This formula directly gives the displacement (a vector quantity).Note that distance (scalar) is equal to displacement only if the motion is in one direction without reversal.Therefore, the area under a velocity-time graph specifically yields net displacement, not average velocity, distance, or average speed.Answer:Option B: its net displacement.