The distance covered by the particle in the l | vedclass

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Question

(B) The distance covered is equal to the area under the velocity-time graph.Total distance covered in $7\,s$ is the area of the trapezium $ABCD$:Area

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The distance covered is equal to the area under the velocity-time graph.Total distance covered in $7\,s$ is the area of the trapezium $ABCD$:Area $= \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Area $= \frac{1}{2} 	imes (6 + 2) 	imes 10 = \frac{1}{2} 	imes 8 	imes 10 = 40\,m$.Distance covered in the last $2\,s$ (from $t = 5\,s$ to $t = 7\,s$) is the area of the triangle $CQD$:Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Area $= \frac{1}{2} 	imes (7 - 5) 	imes 10 = \frac{1}{2} 	imes 2 	imes 10 = 10\,m$.The required fraction is $\frac{10}{40} = 0.25$.

The distance covered by the particle in the last $2\,s$ is what fraction of the total distance covered by it in $7\,s$?

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