1 year ago

Someone help with this fast!

1 answer:

8 0

2x+40=60
2x=20
x=10
Answer:10

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Valentin [98]

Add all the displacement vectors together, then find the magnitude of this vector sum.

$\vec r_1=(72\,\mathrm{km})(\cos60^\circ\,\vec\imath+\sin60^\circ\,\vec\jmath)$

$\vec r_2=(48\,\mathrm{km})(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

$\vec r_3=(100\,\mathrm{km})(\cos150^\circ\,\vec\imath+\sin150^\circ\,vec\jmath)$

Summing these vectors gives the resultant displacement

$\vec r=\vec r_1+\vec r_2+\vec r_3\approx(-50.6\,\vec\imath+64.4\,\vec\jmath)\,\mathrm{km}$

and we have

$\|\vec r\|=\sqrt{(-50.6)^2+64.4^2}\,\mathrm{km}\approx\boxed{81.9\,\mathrm{km}}$

so the plane ends up about 81.9 km away from its starting position.

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uranmaximum [27]

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3 years ago