Physics, 22.01.2020 00:31 krystalScott17

An airplane heads northeast at an airspeed of km/hr, but there is a wind blowing from the west at km/hr. in what direction does the plane end up flying

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The direction of the plain would be northeast east. If a wind is blowing from the west, it goes from west to east. The final speed of the plain is the vectorial addition of the wind speed and the relative speed of the plain. Therefore the flight direction of the plain seen from a stationary point in the earth is obtained from the sum of the direction of the wind and the plain.

Answer from: maskythegamer

Resultant velocity = 711.2km/hr

Direction = 41°

(41° North of east)

Complete Question:

An airplane heads northeast at an airspeed of 660km/hr, but there is a wind blowing from the west at 70 km/hr. In what direction does the plane end up flying?

Explanation:

Given;

Airplane velocity = 660km/hr northeast

Wind blowing velocity = 70 km/hr from the west (towards east)

Finding the horizontal and vertical component of the speed.

For vertical component Ry:

Ry = 660sin45

Ry = 466.7km/hr

For the horizontal component Rx:

Rx = 660cos45 + 70

Rx = 536.7km/hr

The Resultant velocity R can be given as;

R = √(Rx^2 +Ry^2)

Substituting Rx and Ry

R = √(536.7^2 + 466.7^2)

R = 711.2 km/hr

The direction can be derived from:

Tan∅ = opposite/adjacent = Ry/Rx

∅ = taninverse (Ry/Rx)

Substituting Ry and Rx

∅ = taninverse (466.7/536.7)

∅ = 41°

Answer from: Quest

One splits the atoms and molecules the other combines them

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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