Question
A particle moves along a straight line. The velocity-time graph is a triangle: velocity rises linearly from to between and , then drops linearly back to at . Find the total displacement and the average velocity over the interval.
Solution — Step by Step
The displacement is the area under a - graph, regardless of the shape. The graph here is a triangle with base and height . We don’t need calculus — geometry handles it.
Average velocity is total displacement divided by total time, not the average of initial and final speeds.
The total displacement is and the average velocity is .
Why This Works
The area under a velocity-time curve gives displacement because , and integrating sums those slivers. Whether the velocity is constant, linearly rising, or some weird curve, the area rule still applies. We only switch to integrals when the shape isn’t a clean triangle or trapezium.
For motion that doesn’t change direction, average velocity equals the area divided by total time. If the particle had reversed direction, we would have to be careful — areas below the time axis count as negative displacement.
Alternative Method
We could split the motion into two phases. From to , acceleration is and displacement is . From to , the particle decelerates from to over , covering . Total . Same answer, more arithmetic.
The classic trap: students compute average velocity as because the particle starts and ends at rest. That’s the average of the velocities at the endpoints, which is meaningless here. Average velocity always means displacement over time.
Common Mistake
Confusing area under the - graph with the area under an - graph. The - area gives change in velocity, not displacement. JEE Main 2023 had a sneaky version where the graph was labelled ” vs ” but the answer choices were in metres — many students lost marks by assuming displacement.