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Question: For the velocity-time graph shown in figure below the distance covered by the body in last two secon...

For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds

A

12\frac{1}{2}

B

14\frac{1}{4}

C

13\frac{1}{3}

D

23\frac{2}{3}

Answer

14\frac{1}{4}

Explanation

Solution

Distance covered in total 7 seconds

= Area of trapeziumABCD=12(2+6)×10\mathbf{ABCD} = \frac{1}{2}(2 + 6) \times 10= 40 m

Distance covered in last 2 second = area of triangle CDQ

=12×2×10=10\frac{1}{2} \times 2 \times 10 = 10m

So required fraction =1040=14= \frac{10}{40} = \frac{1}{4}