A swimmer heads perpendicular to a 2-miles-per-hour current. His speed in still water is 1.8 miles per hour. Find the actual speed of the swimmer and his direction angle, 0 (the zero right her is supposed to have a line through the middle of it, I think it means cosine or something), with respect to the direction of the current.
see other post.
To find the actual speed of the swimmer and the direction angle, we can use vector addition.
Let's assume the direction of the current is considered as the positive x-axis. The swimmer's velocity can be represented as a vector with magnitude and direction.
Let v_swimmer be the velocity of the swimmer, v_current be the velocity of the current, and v_resultant be the resultant vector of the swimmer's velocity relative to the current.
The magnitude of the swimmer's velocity in still water is given as 1.8 mph, and the speed of the current is 2 mph. We need to find the magnitude of the swimmer's actual velocity.
To find the magnitude of the swimmer's actual velocity, we can use the Pythagorean theorem:
v_swimmer^2 = v_resultant^2 = (v_stillwater)^2 + (v_current)^2
v_swimmer^2 = (1.8 mph)^2 + (2 mph)^2
v_swimmer^2 = 3.24 mph^2 + 4 mph^2
v_swimmer^2 = 7.24 mph^2
Taking the square root of both sides, we find:
v_swimmer = sqrt(7.24 mph^2)
v_swimmer ≈ 2.69 mph
Therefore, the actual speed of the swimmer is approximately 2.69 mph.
To find the direction angle of the swimmer with respect to the direction of the current (θ), we can use trigonometry.
Since the swimmer is moving perpendicular to the current, the direction vector can be considered as the hypotenuse of a right triangle.
We can use the cosine function (cos(θ)) to find this angle:
cos(θ) = adjacent/hypotenuse
cos(θ) = v_current/v_swimmer
cos(θ) = 2 mph / 2.69 mph
cos(θ) ≈ 0.743
To find θ, we can take the inverse cosine (arccos) of 0.743:
θ = arccos(0.743)
θ ≈ 41.21 degrees
Therefore, the direction angle of the swimmer with respect to the direction of the current is approximately 41.21 degrees.