JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Kinematics

Displacement is the signed area under the velocity-time graph. From $0$ to $2$ s the area is $+8\times 2=16$ m, from $2$ to $4$ s it is $-4\times 2=-8$ m, from $4$ to $8$ s it is $+4\times 4=16$ m, and from $8$ to $10$ s it is $-4\times 2=-8$ m, giving a net displacement of $16-8+16-8=16$ m. The distance is the sum of the magnitudes of these areas, $16+8+16+8=48$ m. The ratio of displacement to distance is $\frac{16}{48}=1:3$.