Answer
$r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$
Work Step by Step
$r = (600\hat{i}-400\hat{j}+200\hat{k})\times 10^3~km$
We can find the final position of each component of $r$ separately.
$r_x = r_{x0}+v_{0x}t+\frac{1}{2}a_xt^2$
$r_x = (600\times 10^6~m)+(9500~m/s)(35~min)(60~s/min)+\frac{1}{2}(40~m/s^2)[(35~min)(60~s/min)]^2$
$r_x = 708\times 10^6~m$
There is no component of velocity or acceleration in the y-direction.
$r_y = -400\times 10^6~m$
There is a component of acceleration in the z-direction.
$r_z = z_0+\frac{1}{2}a_zt^2$
$r_z = (200\times 10^6~m)+\frac{1}{2}(-20~m/s^2)[(35~min)(60~s/min)]^2$
$r_z = 156\times 10^6~m$
We can add the components to find the final position of the spaceship.
$r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$
