MHTCET 11 May 2024 Shift 1 Paper

Concept:In a velocity-time graph, the distance covered is equal to the area under the graph.Explanation:The total distance is the area of the trapezium $\mathrm{ABCD}$.Here, the parallel sides are $BC = 2$ and $AD = 6$, and the height is $BF = 10$.So, total distance is:$$\text{Area of }\square ABCD = \frac{1}{2}(BC + AD) \times BF$$$$= \frac{1}{2}(2 + 6) \times 10 = 40\ \text{m}$$Thus, the total distance covered is $40\ \text{m}$.The distance covered in the last two seconds is the area of $\triangle CED$.Its base is $ED = 2$ and its height is $CE = 10$.So,$$S_1 = \text{Area of }\triangle CED = \frac{1}{2} \times ED \times CE$$$$= \frac{1}{2} \times 2 \times 10 = 10\ \text{m}$$Hence, the required ratio is:$$\frac{S_1}{\text{Total distance}} = \frac{10}{40} = \frac{1}{4}$$Answer:$\frac{1}{4}$ Therefore, the correct option is B.