A point initially at rest moves along X-axis. Its acceleration varies with time as a =(6t + 5) ms-2. If it starts from origin, then the distance covered in 2 s is

Given, acceleration, a: =6t+ 5∴              a=dvdt=6t+5,dv=(6t+5)dtIntegrating it, we have                   ∫0vdv=∫0t(6t+5)dtv = 3t2 + 5t + C, where C is a constant of integration.When t = 0, V = 0, so C = 0∴             v=dsdt=3t2+5t   or  ds=3t2+5tdtIntegrating it within the conditions of motion, i. e. as t changes from 0 to 2s, s changes from 0 to s, we have                 ∫0sds=∫023t2+5tdt∴                      s=t3+52t202=8+10=18 m