Answer
The spatial coordinate of the event according to S' is $~~x' = 6.54\times 10^8~m$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{1-\beta^2}$
$\gamma = \frac{1}{1-0.400^2}$
$\gamma = 1.09$
We can find the spatial coordinate $x'$:
$x' = \gamma~(x-vt)$
$x' = 1.09~[3.00\times 10^8~m-(-0.400)(3.0\times 10^8~m/s)(2.50~s)]$
$x' = 1.09~[3.00\times 10^8~m+(0.400)(3.0\times 10^8~m/s)(2.50~s)]$
$x' = (1.09)~(6.00\times 10^8~m)$
$x' = 6.54\times 10^8~m$
The spatial coordinate of the event according to S' is $~~x' = 6.54\times 10^8~m$
