A submarine is travelling parallel to the surface of the ocean at a depth of 626m. It begins a constant ascent in order to reach the surface after travelling a distance of 4420m. The ascent takes 35mins.
A.) what angle of ascent or angle of elevation would the submarine need to make to reach the surface in 4420m.
B.) how far did the submarine travel horizontally during its ascent to the surface?
C.) determine the horizontal speed of the submarine during its ascent?
Please show me all the steps !!! Thanks
draw a diagram.
A) sinθ = 626/4420
B) cotθ = x/626
C) v=x/35 m/min
To solve these questions, we can use trigonometry and physics principles. Let's go step by step.
A.) To find the angle of ascent or angle of elevation, we can use the tangent function. The tangent of an angle is equal to the opposite side divided by the adjacent side. In this case, the opposite side is the depth of the submarine (626m) and the adjacent side is the distance traveled horizontally (4420m).
Let's calculate the angle:
tan(theta) = opposite/adjacent
tan(theta) = 626/4420
You can find the value of theta by taking the inverse tangent (arctan) of both sides:
theta = arctan(626/4420)
Using a scientific calculator or online tool to calculate arctan(626/4420), we find that the angle of ascent is approximately 8.14 degrees.
B.) To find the distance the submarine traveled horizontally during its ascent to the surface, we can use the angle of elevation and the total distance traveled (4420m). We can use the trigonometric function cosine since it relates the adjacent side to the hypotenuse.
Let's calculate the horizontal distance:
cos(theta) = adjacent/hypotenuse
cos(8.14) = adjacent/4420
Rearranging the equation, we can solve for the adjacent side:
adjacent = 4420 * cos(8.14)
Using a scientific calculator or online tool to calculate 4420 * cos(8.14), we find that the submarine traveled approximately 4391.52m horizontally during its ascent.
C.) To determine the horizontal speed of the submarine during its ascent, we can divide the horizontal distance traveled (4391.52m) by the time taken (35 minutes). However, we need to convert the time to seconds because the SI unit of speed is meter per second.
Let's calculate the horizontal speed:
horizontal speed = horizontal distance / time taken
horizontal speed = 4391.52 / (35 * 60) (converting minutes to seconds)
Using a calculator to perform the division, we find that the horizontal speed of the submarine during its ascent is approximately 2.08 m/s.
A.) To determine the angle of ascent or angle of elevation, we can use the tangent of the angle. The formula for tangent is:
tan(angle) = opposite/adjacent
In this scenario, the opposite is the change in depth (626m - 0m = 626m) and the adjacent is the distance traveled horizontally (4420m). Therefore, we have:
tan(angle) = 626/4420
Now we can solve for angle:
angle = atan(626/4420)
Using a calculator, we find:
angle ≈ 7.99 degrees
Therefore, the angle of ascent or angle of elevation would be approximately 7.99 degrees.
B.) The distance traveled horizontally can be found using the cosine of the angle. The formula for cosine is:
cos(angle) = adjacent/hypotenuse
In this scenario, the adjacent is the distance traveled horizontally and the hypotenuse is the total distance covered by the submarine (4420m). Therefore, we have:
cos(angle) = adjacent/4420
Now we can solve for the adjacent (distance traveled horizontally):
adjacent = cos(angle) * 4420
Using the angle calculated in part A (7.99 degrees), we find:
adjacent ≈ cos(7.99) * 4420
adjacent ≈ 4409.489m
Therefore, the submarine traveled approximately 4409.489m horizontally during its ascent to the surface.
C.) The horizontal speed of the submarine during its ascent can be determined by dividing the distance traveled horizontally by the time taken.
horizontal speed = distance traveled horizontally / time taken
Given that the distance traveled horizontally is approximately 4409.489m and the time taken is 35 minutes (or 35 * 60 = 2100 seconds), we have:
horizontal speed = 4409.489 / 2100
horizontal speed ≈ 2.10 m/s
Therefore, the horizontal speed of the submarine during its ascent is approximately 2.10 m/s.