A point initially at rest moves along X-axis. Its acceleration varies with time as a = (6 t + 5) in m/s2. If it starts from origin, the distance covered in 1 sec is

Given that a=(dv /dt) =6t+ 5∴ dv=(6t+5)dtIntegrating, we getV=6t22+5t+C1where C1 is a constant of integration.When t =0,v=0 and hence C1 = 0Now, V=2t2+5t  or ds=3t2+5tdtIntegrating we get s=3t33+5t22+C2When t = 0,s=0 and henceC2 = 0∴s=t3+5t22=1+52 (∵t=1)s = 3.5metre