Graphs of Motion question from MHT CET 2024 (9 May, Shift 1)

Answer: (c)

1. Distance = area under the $v$-$t$ graph from 6 s to 9 s.
2. Slope from 6 s to 10 s: $a = \frac{40 - 20}{4} = 5\text{ m/s}^2$, so $v(9) = 20 + 5\times 3 = 35\text{ m/s}$
3. Area = trapezium $= \frac{1}{2}(20 + 35)\times 3 = 82.5\text{ m}$
4. So the answer is (c) 82.5 m.