An airplane has rest length $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m}}$ and speed $\mathbf{630}\mathbf{}\mathbf{\text{m/s}}$. To a ground observer, (a) by what fraction is its length contracted and (b) how long is needed for its clocks to be $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathbf{\text{s}}$ slow.

The given data can be listed below as:

• The rest length of the airplane is $l=40.0\text{m}$.
• The speed of the airplane is $v=630\text{m/s}$.
• The clock is slow about $t=1.00\mu \text{s}$.

The rest length is described as the object’s length that an observer measures by standing relative to the object. The rest length is measured when the object is at rest.

(a)

The equation of the contraction of length is expressed as:

$L=\frac{v}{2c}$

Here, $L$is the contraction of length,$v$ is the speed of the airplane, and $c$is speed of the light.

Substitute $630\text{m/s}$for $v$and $3\times {10}^8\text{m/s}$for $c$in the above equation.

$$\begin{gathered} L=\frac{630\text{m/s}}{2\left(3\times {10}^8\text{m/s}\right)} \\ =\frac{630\text{m/s}}{\left(6\times {10}^8\text{m/s}\right)} \\ =2.21\times {10}^{-12} \end{gathered}$$

Thus, the fraction of the contracted length is $2.21\times {10}^{-12}$.

(b)

The equation of the time needed for its clocks is expressed as:

$t_1=\frac{2tc}{v}$

Here, $t_1$is the time needed for its clocks and $t$is the slowness of the clock.

Substitute the values in the above equation.

$$\begin{gathered} t_1=\frac{2\left(1\times {10}^{-6}\text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(630\text{m/s}\right)} \\ =\frac{2\left(3\times {10}^2\text{m}\right)}{\left(630\text{m/s}\right)} \\ =\frac{\left(6\times {10}^2\text{m}\right)}{\left(630\text{m/s}\right)} \\ =0.95\text{s} \end{gathered}$$

Thus, the clocks needed a time period of $0.95\text{s}$.