For the acceleration-time (a-t) graph shown i | vedclass

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(B) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.From the graph, the area from $t=0$ to $t=4 \,

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.From the graph, the area from $t=0$ to $t=4 \, s$ is a triangle with base $4 \, s$ and height $4 \, m/s^2$:Area$_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4 = 8 \, m/s$.The area from $t=4 \, s$ to $t=6 \, s$ is a triangle below the time axis with base $(6-4) = 2 \, s$ and height $-4 \, m/s^2$:Area$_2 = \frac{1}{2} 	imes 2 	imes (-4) = -4 \, m/s$.The total change in velocity is the sum of these areas:$\Delta v = 	ext{Area}_1 + 	ext{Area}_2 = 8 + (-4) = 4 \, m/s$.Thus, the correct option is $(b)$.

For the acceleration-time $(a-t)$ graph shown in the figure, the change in velocity of the particle from $t=0$ to $t=6 \, s$ is ........ $m/s$.

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