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3 years ago

8

The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin

g toward the southwest relative to the earth.

1 answer:

leva [86]3 years ago

6 0

<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>

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Explanation:

Let's start writing the vertical position equation :

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Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

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$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

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$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

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The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

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$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

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$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

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$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

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