Physics, 19.05.2021 15:50 lucretiak74

Sarah is demolishing two clock towers, one on either side of the city. She sets up the explosives to go off at the same time when she sends the signal from the center of the city. She sends the signal and sees both clock towers blow up at the same time and sees both towers show 1pm exactly. Adam is flying past the earth traveling at a speed close to the speed of light as sees the building farthest away from him blow up first. Explain how Sarah can see them blow up at the same time, but Adam sees them blow up at different times.

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Answer from: jayy2x41

According to Lorentz transformation, the time observed of by objects moving in different reference frames are related to their velocity of motion and the speed of light

The time measured by Sarah who is located at the source of the signal and the center of the explosion is different from the time measured by Adam who is moving relative to Sarah and observes the signals moving at different speeds and therefore observes the buildings blow up not immediately but at different times

Explanation:

The time and space dilation equations are given as follows;

$\Delta t = \dfrac{\Delta t'}{\sqrt{1 - \dfrac{v^2}{c^2} } }$

$L = L' \times \sqrt{1 - \dfrac{v^2}{c^2} }$

Where;

Δt' = The time measured by Adam

Δt = The time measured by Sarah

L' = The distance measured by Adam

L= The distance measured by Sarah

v = The velocity at which Adam is moving

Let 'u' represent the speed of the signals

Given that Adam is traveling in between the two clock towers such that one signal is moving in the opposite direction towards him while the other signal moves in the his direction, by relativistic velocity addition, we have;

The velocity with which Adam observes the signal coming towards him, v' is given as follows;

v'' = (v + u)/(1 + v·u/c²)

∴ v'' > v

For the signal moving in the direction of Adam, we have;

v''' = (v - u)/(1 - v·u/c²)

v''' < v

Therefore, we have;

The time it takes the light to reach Adam from the furthest Tower which he passes as he is flies past Sarah in the middle, Δt'' is given as follows;

$\Delta t'' = {\Delta t'} \times {\sqrt{1 - \dfrac{v''^2}{c^2} } }$

For the closer tower which Adam approaches, we have;

$\Delta t''' = {\Delta t'} \times {\sqrt{1 - \dfrac{v'''^2}{c^2} } }$

Given that v'' > v''', therefore Δt'' < Δt''' and the time the light from the furthest building is seen first by Adam before the light from the closer building.

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$