Originally posted by voltaire
t'=t0/sqrt(1-(v/c)2)
ok, but I think the equation is
t' = t0/(sqrt(1-(v/c)^2)
you had
t'=t0/sqrt(1-(v/c)2)
the "^" means "to the power of".
anyhow, the above relationship is to velocity, and you have acceleration in your problem.
so by substitution
v = a*(t0)
so
t' = t0/(sqrt(1-((a*t0)/c)^2)
so now time dialation is a function of the initial time.
so the equation looks like
t' = t0/(sqrt(1-[(a/c)^2]*t0^2)
lets let (for the sake of clarity) (a/c)^2 = b
now the equation becomes
t' = t0/(sqrt(1-b*(t0)^2)
now find the rate of change of t' with respect to t0 (ie dt'/dt0)
you'll need to use the product rule, or the quotient rule ( whichever floats your boat)in combination with the chain rule to arrive at
dt'/dt0 = 1/[(1-b*(t0)^2)^(3/2)]
then just evaluate it a t0 = 4(s), as long as everything is in standard units that is the result you desire to obtain.
any questions about the actual differentiation processes, and calculation, ask
later