Question

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.

Short answer

(a) The fraction of the contracted length is $2.21\times {10}^{-12}$.

(b) The clocks needed a time period of $0.95\text{s}$.

Step-by-step solution

Identification of given data

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 $t=1.00\mu \text{s}$.

Significance of the rest length

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.

Determination of the contraction of length

(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 $2.21\times {10}^{-12}$.

Determination of the time needed

(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 $0.95\text{s}$.