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# Vectors and bearings watch

1. So the question is:
Take i = 1 m s^-1 due east and j = 1 m s^-1 due north.

A barge has a speed in still water of 3.4 m s^-1. It is pointed in the direction S32degreesE, but sails in a current of speed 0.73 m s^-1 from the direction S61degreesE.

So far this is my working out,

V(b) - velocity of the barge in still water.

V(b) has magnitude of 3.4. Now it's working out the direction that i'm more uncertain on. I have got, 90 degrees + 32 degrees = 122 degrees, or is it -(90 + 32) = -122 degrees??

V(c) - Velocity of current.

V(c) has a magnitude of 0.73. The direction of V(c) is N61degreesW(because it's from the direction S61degreesE). So would this be 90 - 61 = -29 degrees?

Basically it's the direction that i'm confused about, I will bold te unsure answers. Any help is appreciated, thanks.
2. (Original post by RHCPfan)
So the question is:
Take i = 1 m s^-1 due east and j = 1 m s^-1 due north.

A barge has a speed in still water of 3.4 m s^-1. It is pointed in the direction S32degreesE, but sails in a current of speed 0.73 m s^-1 from the direction S61degreesE.

So far this is my working out,

V(b) - velocity of the barge in still water.

V(b) has magnitude of 3.4. Now it's working out the direction that i'm more uncertain on. I have got, 90 degrees + 32 degrees = 122 degrees, or is it -(90 + 32) = -122 degrees??

V(c) - Velocity of current.

V(c) has a magnitude of 0.73. The direction of V(c) is N61degreesW(because it's from the direction S61degreesE). So would this be 90 - 61 = -29 degrees?

Basically it's the direction that i'm confused about, I will bold te unsure answers. Any help is appreciated, thanks.
Bearings are usually measured clockwise from North, so your two bearings would be 122 and 331 degrees.
3. (Original post by brianeverit)
Bearings are usually measured clockwise from North, so your two bearings would be 122 and 331 degrees.
4. (Original post by RHCPfan)
360-29. !
5. Got ya. Thanks guys.

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Updated: January 23, 2014
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