Question

For the velocity ($v$) -time ($t$) graphs shown in figure, the total distance covered by the particle in the last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds?

• A. 1/8
• B. 1/6
• C. 1/4
• D. 1/2

Answer 1

The correct option is C 1/4

Distance = area under velocity-time graph

$s_{1-3}=\frac{1}{2}\times 2\times 20=20m$ (area of $△$ AEB)

$s_{3-5}=2\times 20=40cm$ (area of rectangle EFCB)

$s_{5-7}=\frac{1}{2}\times 2\times 20=20m$ ( area of $△$ CDF)

$\frac{s_{5-7}}{s_{total}}=\frac{20}{20+40+20}=\frac{20}{90}=\frac{1}{4}$