Kinematics: Edge Cases and Subtle Traps (9)

Question

A particle moves along the x-axis with velocity $v(t) = 6t - t^2$ m/s for $0 \le t \le 8$ s. Find (a) the displacement and (b) the total distance travelled in this interval. This is a classic JEE Main trap because most students treat displacement and distance as the same thing.

Solution — Step by Step

Why This Works

Velocity is a signed quantity. When we integrate $v(t)$ directly, the regions where the particle moves backward subtract from the regions where it moves forward — that gives us net displacement. Distance ignores direction; we always add positive contributions, which is why we integrate $|v(t)|$.

The trap is this — students compute one number and report it as “the answer,” not realising the question wants two. JEE Main 2023 had a near-identical problem worth 4 marks where ~40% of test-takers gave only displacement.

Alternative Method

For polynomial velocity, you can find the position function $x(t) = 3t^2 - t^3/3$, evaluate $x(8) - x(0)$ for displacement, and use turning-point analysis ($|x(6) - x(0)| + |x(8) - x(6)|$) for distance. Same answer, slightly less notation.

Common Mistake

The other subtle error is forgetting the absolute value when computing $d_2$. Students write $d_2 = \int_6^8 (6t - t^2)\, dt = -\tfrac{44}{3}$ and either drop the negative sign mid-calculation or add the negative number, getting distance $<$ displacement. Distance can never be less than displacement — that itself is a sanity check worth doing.