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Q. For the acceleration-time (a-t) graph shown in figure, the change in velocity of particle from $t=0$ to $t=6\, s$ isPhysics Question Image

Motion in a Straight Line

Solution:

Area under $a-t$ graph gives change in velocity.
So, $\Delta v=\frac{1}{2} \times 4 \times 4-\frac{1}{2} \times 2 \times 4=8-4$
$\Delta v=4\, ms ^{-1}$

Solution Image