MHTCET 13 Apr 2026 Shift 2 Paper

Concept:The area under a velocity-time graph gives displacement, while the total area without sign gives distance travelled.Explanation:From the graph, divide the area into four regions from $t = 0$ to $t = 10$ s.Region 1 (0 to 2 s): rectangle of height $8$ and width $2$$A_1 = 8 \times 2 = 16$ m (above axis, positive)Region 2 (2 to 4 s): rectangle of height $-4$ and width $2$$A_2 = -4 \times 2 = -8$ m (below axis, negative)Region 3 (4 to 8 s): rectangle of height $4$ and width $4$$A_3 = 4 \times 4 = 16$ m (above axis, positive)Region 4 (8 to 10 s): rectangle of height $-4$ and width $2$$A_4 = -4 \times 2 = -8$ m (below axis, negative)Displacement = sum of signed areas:$16 - 8 + 16 - 8 = 16$ mDistance = sum of magnitudes of areas:$16 + 8 + 16 + 8 = 48$ mRequired ratio:$\text{Displacement} : \text{Distance} = 16 : 48 = 1 : 3$Answer:D. $1 : 3$