Physics, 14.09.2019 07:30 ortizprecious5183

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second oasis. what distance separates the two oases?

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Answer from: austinwst3

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

OB^2 = BC^2 + OC^2

PUTTING ALL VALUE IN ABOVE EQUATION

d^2 = 930-25sin30)^2 + (25cos30)^2

d^2 = 775

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km t

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.

Answer from: Quest

The boat's velocity relative to the riverbank can be resolved into two components. the component going north is 2.7 m/s and the component going west is 1.2 m/s.we add these two vectors and then find the magnitude of the resultant to get the speed.the magnitude of the resultant vector from adding can be obtained by the pythagorean theorem. $\begin{aligned} | \vec{v}| & = \sqrt{(\text{2.7 m/s})^2 + (\text{1.2 m/s})^2 } \\ & = 2.954 \text{ m/s} \end{aligned}$ to two significant figures, the boat's speed relative to the riverbank is 3.0 m/s.

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