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Question

(B) The distance travelled by a particle is equal to the area under the velocity-time $(v-t)$ graph.From the given graph,the total distance is the sum

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) The distance travelled by a particle is equal to the area under the velocity-time $(v-t)$ graph.From the given graph,the total distance is the sum of the areas of the four geometric shapes $(A_1, A_2, A_3, A_4)$:$1$. Area $A_1$ (triangle from $t=0$ to $t=1$): $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 20 = 10 \; m$.$2$. Area $A_2$ (rectangle from $t=1$ to $t=2$): $	ext{length} 	imes 	ext{breadth} = 1 	imes 20 = 20 \; m$.$3$. Area $A_3$ (trapezium from $t=2$ to $t=3$): $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height} = \frac{1}{2} 	imes (20 + 10) 	imes 1 = 15 \; m$.$4$. Area $A_4$ (rectangle from $t=3$ to $t=4$): $	ext{length} 	imes 	ext{breadth} = 1 	imes 10 = 10 \; m$.Total distance = $A_1 + A_2 + A_3 + A_4 = 10 + 20 + 15 + 10 = 55 \; m$.

The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in $4 \; s$ is ......... $m$.

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