Chapter 15, Problem 35A

Thomas, a rhino, is 2.5 meters long when at rest.

a. How long will you measure him to be when he’s running by at 0.80c?

b. Show that you would measure a time of 6.3 ns for the length of his body to pass you.

To determine

The length of the person measured by an observer.

Answer to Problem 35A

The length of the person measured by an observer is $1.5\text{ m}$ .

Explanation of Solution

Given:

The length of a rhino is $L_0=2.5\text{ m}$ .

The speed of rhino is $v=0.8c$ .

Formula used:

The expression for relativistic length contraction is,

  $L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Here, $L_0$ is the measured length of the object at rest, $v$ is the speed of the object relative to an observer, and $c$ is the speed of the light.

Calculation:

The length of the person measured by an observer is,

  $\begin{array}{l}L=L_0\sqrt{1-\frac{v^2}{c^2}}\\ L=\left(2.5\text{ m}\right)\sqrt{1-\frac{{\left(0.8c\right)}^2}{c^2}}\\ L=1.5\text{ m}\end{array}$

Conclusion:

Thus, the length of the person measured by an observer is $1.5\text{ m}$ .

To determine

To show: The time taken for the length of body to cross the person is $6.3\text{ ns}$

Explanation of Solution

Given:

The length of a rhino is $L_0=2.5\text{ m}$ .

The speed of rhino is $v=0.8c$ .

Formula used:

The expression for time taken is,

  $t=\frac{L}{v}$

Here, $L$ is the length of the moving object and $v$ is the speed of the object relative to an observer.

Calculation:

From part (a), the length of the person measured by an observer is $L=1.5\text{ m}$ .

Consider the speed of light as $c=3\times {10}^8\text{m}/\text{s}$ .

The time taken for the length of body to cross the person is,

  $\begin{array}{l}t=\frac{L}{v}\\ t=\frac{1.5\text{m}}{0.8c}\\ t=\frac{1.5\text{m}}{0.8\times \left(3\times {10}^8\text{m}/\text{s}\right)}\\ t=6.3\times {10}^{-9}\text{ s}\\ t=6.3\text{ ns}\end{array}$

Conclusion:

Thus, the time taken for the length of body to cross the person is $6.3\text{ ns}$ .