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Chapter 33: Problem 36

Two spaceships are racing. The "slower" one passes Earth at \(0.70 c,\) and the "faster" one moves at \(0.40 c\) relative to the slower one. What's the faster ship's speed relative to Earth?

Short Answer

Expert verified
The speed of the faster spaceship, relative to Earth, is \(0.84c\).

Step by step solution

01

Identify the given information

Here, the speeds of the two spaceships are given. The slower one is moving at a speed of \(0.70c\) relative to the Earth, where \(c\) is the speed of light, and the faster one is moving at a speed of \(0.40c\) relative to the slower spaceship.
02

Apply the formula for velocity addition

Using the formula from special relativity for the addition of velocities, \(v = \frac{v1 + v2}{1 + \frac{v1v2}{c^2}}\), where \(v1\) is the speed of the slower spaceship relative to the Earth, \(v2\) is the speed of the faster spaceship relative to the slower one, and \(v\) is the speed of the faster spaceship relative to the Earth. Substituting the given values, \(v = \frac{0.70c + 0.40c}{1 + \frac{(0.70c)(0.40c)}{c^2}}\).
03

Simplify the equation

Solving the equation, we get \(v = \frac{1.10c}{1 + 0.28}\).
04

Calculate the final answer

Finally, solving for \(v\), we get \(v = 0.84c\). This is the speed of the faster spaceship relative to the Earth.

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