Kinematics Problem with Rockets

[Minestra]

Homework Statement

In Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A better expression for a rocket's position measured from the center of the Earth is given by
Y(t)=(R_E^3/2+3(√g/2)R_Et)^2/3
where R_E is the radius of the Earth (6.38 ✕ 106 m) and g is the constant acceleration of an object in free fall near the Earth's surface (9.81 m/s2).

Find V_y(t) and a_y(t) using variables not numerical values.

Then what is t when y=4R_E
Lastly find V and a at t = y @ 4R_E2. The attempt at a solution
Naturally I found the first and second derivative of Y(t), which were not accepted by webassign (not correct nor incorrect). Then being stumped by that I decided to find t when y = 4R_E, which also gave me an incorrect answer.

I got Y'(t)=2/3(R_E^3/2+3√(g/2)R_Et)^(-1/3)(3√(g/2)R_E
Y"(t)=-2/9(R_E^3/2+3√(g/2)R_Et)^(-4/3)(3√(g/2)R_E

I also found a t value of 9.5947 when y = 4R_E, this was found by just plugging in all the values in Y(t).

I know I just posted a problem a day or to ago, but I've been beaten by this one aswell. Any advice on how to go about this would be a great help. Thank you.

 

[gneill]

In your Y(t) equation, is the 'g' term: $\frac{\sqrt{g}}{2}$ or $\sqrt{\frac{g}{2}}$ ?

 

[Minestra]

In your Y(t) equation, is the 'g' term: $\frac{\sqrt{g}}{2}$ or $\sqrt{\frac{g}{2}}$ ?

$\sqrt{\frac{g}{2}}$

Sorry I'm still figuring out formatting on here.

 

[gneill]

Your first derivative looks okay to me. Unfortunately. with complex expressions like this there are usually several ways to "simplify" it, and webassign may not recognize the one you've reached. It looks to me that the 3's in your expression could cancel, for example.

I can't seem to make your second derivative match what I find. So maybe you'll need to revisit that.

You'll have to show how you arrived at t = 9.5947 s for y = 4R_E. When I plug in that value for t I get just over 1 R_E.