Q.
Acceleration-time graph for a particle is given in figure. If it starts motion at $t=0$, distance travelled in $3\, s$ will be
Motion in a Straight Line
Solution:
Draw the $v-t$ graph from $a-t$ graph.
Area under $v-t$ graph $=\frac{1}{2} \times 2 \times(3+1)$
$=4\, m$
