A ship sails 15.0 mi on a course S40°10’W and then 21.0 mi on a course N28°20’W. Find the distance and direction of the last position from the first.
I wanna know
To find the distance and direction of the last position from the first, we can use vector addition.
1. Start by converting the given courses S40°10’W and N28°20’W into their respective components.
- The course S40°10’W implies a bearing of 220° (180° + 40°) with respect to the positive x-axis.
- The course N28°20’W implies a bearing of 321° (360° - 28° - 20°) with respect to the positive x-axis.
2. Convert the distances 15.0 mi and 21.0 mi into their respective components.
- Since the first course is in the southwestern direction,
the x-component of the first displacement would be: 15.0 mi * sin(220°) = -13.86 mi
the y-component of the first displacement would be: 15.0 mi * cos(220°) = -10.46 mi
- Since the second course is in the northwestern direction,
the x-component of the second displacement would be: 21.0 mi * sin(321°) = -14.71 mi
the y-component of the second displacement would be: 21.0 mi * cos(321°) = 11.87 mi
3. Add the x-components and y-components separately to get the net displacement.
- Net x-component: (-13.86 mi) + (-14.71 mi) = -28.57 mi
- Net y-component: (-10.46 mi) + (11.87 mi) = 1.41 mi
4. The distance between the first and last position is the magnitude of the net displacement.
- Distance = √[(-28.57 mi)^2 + (1.41 mi)^2] = √(817.04 mi^2) = 28.61 mi
5. The direction can be calculated by finding the angle made by the net displacement with respect to the positive x-axis.
- Direction = arctan(1.41 mi / -28.57 mi) = -2.78°
Therefore, the distance of the last position from the first is approximately 28.61 miles in a direction of S2°48’W.
how did you get these (21,118°20')?
if we indicate a path by (r,Ø) , where r is the distance and Ø is the direction measured counterclockwise from 0 at the normal position
then in component form (rcosØ,rsinØ we have
(15,S W40°10' W) = (15,229°50') = (-9.6752, -11.46257)
(21, N 28°20') = (21,118°20') = -9.9666, 18.4842)
add those two vectors for ( -19.6428, 7.021658)
magnitude = √(-19.6428^2+7.02168^2) = 20.859
direction:
tanØ = .7021.../-19.642.. = ..
Ø = W 19°40' N
check my arithmetic