For the velocity-time graph shown in the figu | vedclass

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(B) The distance covered by an object in a velocity-time graph is equal to the area under the graph.$1$. Total distance covered in $7 	ext{ s}$ (Area

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The distance covered by an object in a velocity-time graph is equal to the area under the graph.$1$. Total distance covered in $7 	ext{ s}$ (Area of the trapezium):$	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$	ext{Area} = \frac{1}{2} 	imes (7 - 1 + 5 - 3) 	imes 10 = \frac{1}{2} 	imes (6 + 2) 	imes 10 = \frac{1}{2} 	imes 8 	imes 10 = 40 	ext{ m}$.$2$. Distance covered in the last $2 	ext{ s}$ (from $t = 5 	ext{ s}$ to $t = 7 	ext{ s}$):This is the area of the triangle formed between $t = 5$ and $t = 7$.$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (7 - 5) 	imes 10 = \frac{1}{2} 	imes 2 	imes 10 = 10 	ext{ m}$.$3$. Required fraction:$	ext{Fraction} = \frac{	ext{Distance in last } 2 	ext{ s}}{	ext{Total distance}} = \frac{10}{40} = 0.25$.

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

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