$$
Delta t_{moving} = frac{Delta t_{rest}}{sqrt{1-frac{v^2}{c^2}}}.
$$

$$
l_{moving} = l_{rest}sqrt{1-frac{v^2}{c^2}}
$$

the length of a moving object appears to you to be less than that of an identical object that rests in respect to you.

These phenomenons are not really observable in everyday life; for example, if the moving observer is in a moving train, he will not experience time dilation and length contraction, because the train is moving at nonrelativistic speeds. However, if he was to find himself in a spaceship, he would experience these phenomenons, because the speed of the spaceship can be compared to the speed of light in vacuum $c$.

Time and length do not have absolute values; those values depend on the frame of reference. But, they can only be experienced when an observer is moving with speeds close to the speed of light.

$t_o = 25$ s

$v = 0.750 c$

$textbf{Find:}$ time, $Delta t$ it takes for the unstable isotope to decay at speed, $v$

Using the time dilation formula we can compute for $Delta t$ as follows:

$$
begin{align*}
Delta t &= dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}} \
&= dfrac{25 text{ s}}{sqrt{1 – dfrac{(0.750 c)^2}{c^2}}} \
&= boxed{37.8 text{ s}}
end{align*}
$$

$L_o = 1.00$ m

$v = 0.910 c$

$textbf{Find:}$ Length, $L$ of the meter stick at speed, $v$

Using the length contraction formula we can compute for $L$ as follows:

$$
begin{align*}
L &= L_o times sqrt{1 – dfrac{v^2}{c^2}} \
&= 1.00 text{ m} times sqrt{1 – dfrac{(0.910 c)^2}{c^2}} \
&= boxed{0.415 text{ m}}
end{align*}
$$

$t = 13.2$ s

$v = 0.885 c$

$textbf{Find:}$ time, $Delta t_o$ elapsed for an observer at rest

Using the time dilation formula we can compute for $Delta t_o$ as follows:

$$
begin{align*}
Delta t_o &= Delta t times sqrt{1 – dfrac{v^2}{c^2}} \
&= 13.2 text{ s} times sqrt{1 – dfrac{(0.885 c)^2}{c^2}} \
&= boxed{6.15 text{ s}}
end{align*}
$$

$L_o = 1.00$ m

$L = 0.950$ m

$textbf{Find:}$ speed of the spaceship, v

We use the length contraction formula to obtain an equation that will help us calculate v, such that:

$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
dfrac{L}{L_o} &= sqrt{1 – dfrac{v^2}{c^2}} \
dfrac{L^2}{L_{o}^2} &= 1 – dfrac{v^2}{c^2} \
dfrac{v^2}{c^2} &= 1 – dfrac{L^2}{L_{o}^{2}} \
v &= c left(1 – dfrac{(0.950 text{ m})^2}{(1.00 text{ m})^2}right) \
v &= 0.0975 c
end{align*}
$$

The speed of the spaceship is $boxed{text{0.0975 c}}$