Two airplanes leave an airport at the same time. The velocity of the first airplane is 650 m/h at a heading of 60.5 ◦ . The velo
city of the second is 560 m/h at a heading of 85◦ . How far apart are they after 2.9 h? Answer in units of m.
1 answer:
First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
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The velocity of the combination of Jackie and the bicycle is 3.328 m/s.
Explanation:
From the given data the constant kinetic energy is 3.6 J. The mass of combination is 0.65 kg. To find the velocity of the combination of Jackie and the bicycle the formula is
KE = 0.5 x mv2.
To find velocity,
V2=ke/(0.5×m)
V=
v= 3.6/(0.5×0.65)
=
v= 3.328 m/s
Hence, the velocity of the combination of Jackie and the bicycle is 3.328m/s.