Answer
(a) The time interval measured by the passenger is $7.5\times 10^{-6}~s$
(b) The time interval measured by an observer on the ground is $1.25\times 10^{-5}~s$
Work Step by Step
(a) Let $L_0 = 3.0~km$
We can find the length $L$ according to the passenger on the train:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$L = (3.0~km)~\sqrt{1-\frac{(0.80~c)^2}{c^2}}$
$L = 1.8~km$
We can find the time interval $\Delta t_0$:
$\Delta t_0 = \frac{1800~m}{(0.80)(3.0\times 10^8~m/s)}$
$\Delta t_0 = 7.5\times 10^{-6}~s$
The time interval measured by the passenger is $7.5\times 10^{-6}~s$
(b) We can find the time interval $t_0$ an observer on the ground measures:
$t_0 = \frac{3000~m}{(0.80)(3.0\times 10^8~m/s)}$
$t_0 = 1.25\times 10^{-5}~s$
The time interval measured by an observer on the ground is $1.25\times 10^{-5}~s$
