From the v - t graph shown, the ratio of distance to displacement in 25 s of motion is: (a) 1 (b) 3/5 (c) 1/2 (d) 5/3

Answer: (d). Areas: 0-5 s: 25; 5-10 s: 50; 10-15 s: 1/2(10+20)× 5 = 75; 15-20 s: 1/2× 5× 20 = 50 (all positive). 20-25 s: 1/2× 5× 20 = 50 below the axis (negative). Distance = 25 + 50 + 75 + 50 + 50 = 250 m; displacement = 200 - 50 = 150 m Ratio = (250)/(150) = 5/3. So the answer is (d).

From the v - t graph shown, the ratio of distance to displacement in… | JEE Main 2023 (11 Apr, Shift 1)

Q16JEE Main 2023 · 11 APR · SHIFT 1Easy

Graphs of Motion

From the $v - t$ graph shown, the ratio of distance to displacement in $25\text{ s}$ of motion is:

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Answer: (d)

Areas: 0-5 s: $25$ ; 5-10 s: $50$ ; 10-15 s: $\frac{1}{2}(10+20)\times 5 = 75$ ; 15-20 s: $\frac{1}{2}\times 5\times 20 = 50$ (all positive).
20-25 s: $\frac{1}{2}\times 5\times 20 = 50$ below the axis (negative).
Distance $= 25 + 50 + 75 + 50 + 50 = 250\text{ m}$ ; displacement $= 200 - 50 = 150\text{ m}$
Ratio $= \frac{250}{150} = \frac{5}{3}$ . So the answer is (d).

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