Physics, 21.01.2020 13:31 wellllm

What is frame of reference equations and how to apply?

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A frame of reference in physics means where the coordinate system (x-y-z axes) are located as a function of time. From the observers perspective at the origin (frame of reference), he can write equations describing the moving objects position, velocity and acceleration as a function of time. Here's an example for you:

Try to imagine two different observers, one observer is on a moving train at 30mph, the other observer is on the ground watching. The observer on the train throws a ball at 20mph, the length of the train car. As far as the coordinate system where the observer is standing, the ball is moving at 20mph away from him. But as far as the observer on the ground sees, the observer on the train is moving 30 mph away from him, and the ball is moving at 30 + 20 = 50 mph away from him. The equations that each observer would write to describe the moving ball would be different because of the different Reference Frames (Coordinate Axes) where the observers are a function of time.

Answer from: Quest

4. 50x10= 500m/s 5. 52m/s
First to answer will be the brainliest answer both questions 4 and 5 and don't answer if you don't k

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

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

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

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

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