Q.
For the acceleration-time (a-t) graph shown in figure, the change in velocity of particle from $t=0$ to $t=6\, s$ is
Motion in a Straight Line
Solution:
Area under $a-t$ graph gives change in velocity.
So, $\Delta v=\frac{1}{2} \times 4 \times 4-\frac{1}{2} \times 2 \times 4=8-4$
$\Delta v=4\, ms ^{-1}$
