91影视

Given data:

The distance traveled by muon observed in a laboratory is d = 800 m .

The actual lifetime of the muon is $t=2脳{10}^{鈥�6}s$.

Let v be the actual speed of the muon and $t^{\text{'}}$be the lifetime observed in a laboratory. Hence,

$t^{\text{'}}=纬t$

${\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ 鈥︹�� (1)

Here,$t^{\text{'}}=\frac{d}{v}$.

Hence, re-write the equation (1),

$\frac{d}{v}=\frac{t}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

Squaring on both sides,

$$\begin{gathered} {\left(\frac{d}{v}\right)}^2=\frac{t^2}{1-{\displaystyle \frac{v^2}{c^2}}} \\ \left(\frac{t^2}{d^2}\right)v^2=1-\frac{v^2}{c^2} \\ {v}^2\left[{\left(\frac{t}{d}\right)}^2+\frac{1}{c^2}\right]=1 \\ \frac{v^2}{c^2}=\frac{1}{1+{\left({\displaystyle \frac{tc}{d}}\right)}^2} \end{gathered}$$ 鈥︹�� (2)

Solve for the value of$\frac{tc}{d}$.

$$\begin{gathered} \frac{tc}{d}=\frac{\left(2脳{10}^{-6}s\right)\left(3脳{10}^{8}m/s\right)}{800m} \\ \frac{tc}{d}=\frac{6脳{10}^2}{800} \\ \frac{tc}{d}=\frac{6}{8} \\ \frac{tc}{d}=\frac{3}{4} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\mathrm{tc}}{\mathrm{d}}\mathrm{in}\mathrm{equation}\left(2\right). \\ \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}=\frac{1}{1+{\left({\displaystyle \frac{3}{4}}\right)}^2} \\ \frac{\mathrm{v}}{\mathrm{c}}=\frac{1}{1+{\displaystyle \frac{9}{16}}} \\ \frac{\mathrm{v}}{\mathrm{c}}=\frac{16}{25} \\ \mathrm{v}=\frac{4}{5}\mathrm{c} \\ \mathrm{On}\mathrm{further}\mathrm{solving}, \\ \mathrm{v}=\frac{4}{5}脳\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right) \\ \mathrm{v}=2.4脳{10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the student has not taken into account the time dilation of the muon鈥檚 鈥渋nternal clock.鈥�

Therefore, the student has not considered the time dilation of the muon鈥檚 鈥渋nternal clock,鈥� and the actual speed of the muon will be $2.4脳{10}^8\mathrm{m}/\mathrm{s}$.