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(A) The distance covered by an object is equal to the area under the velocity-time graph.To find the distance between $t = 0 \; s$ and $t = 4 \; s$,we

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$30$

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(A) The distance covered by an object is equal to the area under the velocity-time graph.To find the distance between $t = 0 \; s$ and $t = 4 \; s$,we calculate the area of the region under the graph from $t = 0$ to $t = 4$.This region consists of a triangle from $t = 0$ to $t = 2$ and a rectangle from $t = 2$ to $t = 4$.Area of triangle (from $t = 0$ to $t = 2$) $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 \; s 	imes 10 \; m/s = 10 \; m$.Area of rectangle (from $t = 2$ to $t = 4$) $= 	ext{length} 	imes 	ext{width} = (4 - 2) \; s 	imes 10 \; m/s = 2 \; s 	imes 10 \; m/s = 20 \; m$.Total distance $= 10 \; m + 20 \; m = 30 \; m$.

The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance (in $m$) covered by the object between $t = 0$ and $t = 4 \; s$?

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