A lift is going up. The total mass of the lif | vedclass

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(D) The distance traveled by the lift is equal to the area under the velocity-time graph.The graph is a trapezium with parallel sides of lengths $12\,

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) The distance traveled by the lift is equal to the area under the velocity-time graph.The graph is a trapezium with parallel sides of lengths $12\, s$ and $(10 - 2) = 8\, s$,and a height of $3.6\, m/s$.Alternatively,the area can be calculated as the sum of the areas of two triangles and one rectangle:Area $= (	ext{Area of first triangle}) + (	ext{Area of rectangle}) + (	ext{Area of second triangle})$Area $= (\frac{1}{2} 	imes 2 	imes 3.6) + (8 	imes 3.6) + (\frac{1}{2} 	imes 2 	imes 3.6)$Area $= 3.6 + 28.8 + 3.6 = 36\, m$.Thus,the total height reached by the lift is $36\, m$.

$A$ lift is going up. The total mass of the lift and the passenger is $1500\, kg$. The variation in the speed of the lift is as given in the graph. The height to which the lift takes the passenger is ............ $m$.

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