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Chapter 5: Problem 32

How fast would a 6.0 m-long sports car have to be going past you in order for it to appear only \(5.5 \mathrm{m}\) long?

Short Answer

Expert verified
The sports car would have to be going at approximately \(1.17\times 10^8 m/s\) for it to appear only 5.5 meters long.

Step by step solution

01

Understand Lorentz contraction formula

Lorentz contraction formula connects the length of an object in the stationary frame (\(L_0\)) with the length observed in a moving frame (\(L\)), given the object's relative speed (\(v\)) and the speed of light (\(c\)): \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \] In our case, \(L_0 = 6.0m\) and \(L = 5.5m\). Our task is to find the required speed \(v\).
02

Rearrange the formula to solve for v

First, we need to rearrange the Lorentz contraction formula to solve for v. Divide both sides of the equation by \(L_0\): \[ \frac{L}{L_0} = \sqrt{1 - \frac{v^2}{c^2}} \] Now, square both sides of the equation: \[ \left(\frac{L}{L_0}\right)^2 = 1 - \frac{v^2}{c^2} \] Next, move the \(v^2/c^2\) term to the left side and the \(\left(\frac{L}{L_0}\right)^2\) term to the right side: \[ \frac{v^2}{c^2} = 1 - \left(\frac{L}{L_0}\right)^2 \] Finally, multiply both sides by \(c^2\) and take the square root of both sides to solve for v: \[ v = c \sqrt{1 - \left(\frac{L}{L_0}\right)^2} \]
03

Plug in the values and compute the speed

Now, let's plug in the values of \(L\), \(L_0\), and \(c\) to find the required speed \(v\). The speed of light is approximately \(c = 3.0 \times 10^8 m/s\). So, the equation becomes: \[ v = (3.0 \times 10^8 m/s) \sqrt{1 - \left(\frac{5.5m}{6.0m}\right)^2} \] Calculate the expression inside the square root: \[ 1 - \left(\frac{5.5m}{6.0m}\right)^2 \approx 0.153 \] Now, multiply by the speed of light and take the square root: \[ v \approx (3.0 \times 10^8 m/s) \sqrt{0.153} \approx 1.17\times 10^8 m/s \] So, the sports car would have to be going at approximately \(1.17\times 10^8 m/s\) for it to appear only 5.5 meters long.

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