Thank you for reporting, we will resolve it shortly
Q.
A particle starts from rest. Its acceleration $(a)$ versus time $(t)$ is as shown in the figure. The maximum speed of the particle will be
Motion in a Straight Line
Solution:
The area under acceleration-time graph gives change in velocity. As acceleration is zero
at the end of $11\, \sec,$ we have
$v_{max} = $ Area of $\Delta OAB$
$ = \frac{1}{2} \times 11 \times 10 = 55\,m/s$
