Astar is transited by a planet. from the measured period t and the transit duration tdur alone, show that one can obtain an upper bound on the density of the transited star: p.,max 3r/(gr2) hint: combine kepler's law and eq. 6.7 to eliminate the separation a, and then extract the density of the spherical star. the upper bound is obtained by assuming an impact parameter b 0. a. b. suppose that from the transit light curve one can measure the total transit duration, tt, between the "first contact" and the "last contact" of the planet and the star, and also the duration of the "flat" part of the transit, t, during which the planet is fully superimposed on the star's disk. show that in this case, utilizing also the transit depth (eq. 6.6), b can be determined, and hence the star's density (and not just upper bound) can be found.

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

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

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.