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>
Answer:
0.84 m
Explanation:
Given in the y direction:
Δy = 0.60 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.35 s
Given in the x direction:
v₀ = 2.4 m/s
a = 0 m/s²
t = 0.35 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²
Δx = 0.84 m
Answer:
The answer are given above in attachment.
The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect
Answer:
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Explanation: