Thank you for reporting, we will resolve it shortly
Q.
The $v-t$ curve shown above is a straight line parallel to time-axis. The displacement in the time interval $t =0$ and $t=T$ is equal to
Motion in a Straight Line
Solution:
For the graph shown, area under the $v-t$ curve represents area of the rectangle of height $\mu$ and base $T$.
$\because$ Area under the $v -t$ curve
$=$ Displacement during $t=0$ and $t=T$
$=\mu \times T =\mu T $