Question
A particle moves in a straight line with velocity (in m/s). Find the displacement of the particle from to s, the distance covered in the same interval, and the time at which the particle momentarily comes to rest. This is a classic JEE Main pattern question — appeared in shifts of 2023 and 2024.
Solution — Step by Step
Set : , giving or s. The particle is at rest at s (the non-trivial root).
Displacement is the signed integral of velocity:
The particle returns to its starting point. Net displacement is zero.
Distance is the unsigned path length, so we split at s where velocity changes sign.
From to : m (positive direction).
From to : m, so distance m.
Total distance m.
Displacement = 0 m, Distance = 8 m, Particle at rest at t = 2 s.
Why This Works
The trick the examiner tests here is the difference between displacement (signed area under v-t curve) and distance (unsigned path length). When velocity changes sign mid-interval, integrating directly gives displacement — but distance needs the integral split at every zero of .
Most students forget Step 3 and report “distance = 0 m”, which is the wrong answer in the option grid. The exam examiner places this trap deliberately because the integral evaluates to a clean zero.
Alternative Method
Sketch as a parabola opening downward, crossing the t-axis at and . The area above the axis from to equals the area below from to , both 4 m² in magnitude. This visual method is faster in MCQs.
A common mistake is treating as distance. It gives displacement. For distance, you must split the integral at every sign change of and add absolute values.
In any v-t numerical, immediately ask: does change sign in the given interval? If yes, distance and displacement will differ. This 5-second check saves the question.