*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