Physics, 11.02.2021 18:30 ryliepeloquinf

g In an historical movie, two knights on horseback start from rest 84.1 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.316 m/s2, while Sir Alfred's has a magnitude of 0.289 m/s2. Relative to Sir George's starting point, where do the knights collide?

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Answer from: Jasoncookies23

The knights will collide at 43.854 meters relative to Sir George's starting point.

Explanation:

Let suppose that initial positions of Sir George and Sir Alfred are 0 and 84.1 meters, respectively. If both knights accelerate uniformly, then we have the following kinematic formulas:

Sir George

$x_{G} = x_{G,o}+v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2}$ (1)

Sir Alfred

$x_{A} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}$ (2)

Where:

$x_{G,o}$ , $x_{A,o }$ - Initial position of Sir George and Sir Alfred, measured in meters.

$x_{G}$ , $x_{A}$ - Final position of Sir George and Sir Alfred, measured in meters.

$v_{o,G}$ , $v_{o,A}$ - Initial velocity of Sir George and Sir Alfred, measured in meters per second.

- Time, measured in seconds.

$a_{G}$ , $a_{A}$ - Acceleration of Sir George and Sir Alfred, measured in meters per square second.

Both knights collide when $x_{G} = x_{A}$ , then we simplify this system of equations below:

$x_{G,o} + v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}$

$(x_{A,o}-x_{G,o}) +(v_{o,A}-v_{o,G})\cdot t +\frac{1}{2}\cdot (a_{A}-a_{G})\cdot t^{2} = 0$ (3)

If we know that $x_{A,o} = 84.1\,m$ , $x_{G,o} = 0\,m$ , $v_{o,A} = 0\,\frac{m}{s}$ , $v_{o,G} = 0\,\frac{m}{s}$ , $a_{A} = -0.289\,\frac{m}{s^{2}}$ and $a_{G} = 0.316\,\frac{m}{s^{2}}$ , then we have the following formula:

$84.1 -0.303\cdot t^{2} = 0$ (4)

The time associated with collision is:

$t \approx 16.660\,s$

And the point of collision is:

$x_{G} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (16.660\,s)+ \frac{1}{2}\cdot \left(0.316\,\frac{m}{s^{2}} \right) \cdot (16.660\,s)^{2}$

$x_{G} = 43.854\,m$

The knights will collide at 43.854 meters relative to Sir George's starting point.

Answer from: Quest

Clike bone scan for find heart

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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