Question
a particle initially at rest moves along x-axis. Its acceleration is given as a = (2t+2) m/s2. If it starts from origin , find distance covered in 2 second.
Joshi sir comment
dv/dt = 2t+2
so dv = 2tdt+2dt
on integrating the given eq.
v = t2 + 2t + C
at start v = 0 at t = 0
so on puting these initial values we get C = 0
so v = t2 + 2t
so dx/dt = t2 + 2t
so dx = t2dt + 2tdt
on integrating x = t3/3 + t2+ C
on putting initial values x=0 at t=0
we get C = 0
so x = t3/3 + t2
now put t = 2