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3 years ago

A jet airliner moving initially at 889 mph

Physics

1 answer:

Eduardwww [97]3 years ago

8 0

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

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$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

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$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

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