dv/dt = g(1+v)
dv/(1+v) = g dt
ln(1+v) = gt
1+v = e^(gt) + c
v = e^(gt) + c (diferent c)
v(0) = 0, so c = -1
v = e^(gt)-1
now, v = ds/dt, so
s = 1/g e^(gt) - t + c
now, e^(gt) = v-1, so
s = 1/g (v-1) - t + c
since v = e^(gt)-1,
ln(v)+1 = gt
t = 1/g (ln(v)+1))
and so
s = 1/g (v-1) - 1/g (ln(v)+1) + c
s = 1/g (v-1-ln(v)-1) + c
You may have to back up a step or two to get things to line up with your proposed answer. But you see how to get there, right? And better double-check my on-the-fly math.
A body is initially at rest at an origin O it subsequently move in a straight line from O with acceleration at an instance given by g(1+v) where v is the velocity of the particle at that instant and g is constant,if S is a displacement of the particle from O at any instance show s=1/g(v-lin 1+v+1/g(1+c)
2 answers
I already see one error:
gt = ln(v+1)
gt = ln(v+1)
2 answers