A skiff leaves a dock and heads toward a house across the river. The house is at a bearing of N 64� E from the dock. There is a 1 mile per hour current blowing due east. Determine the speed and direction the skiff would have to maintain so that the skiff’s actual speed is 4 miles per hour and it is moving directly toward the house. Round the speed to the nearest whole number and the direction to the nearest degree.

Guest May 4, 2015

#2**+10 **

We can solve this with a "tip-to-tail" method.

Let the "flowing water " vector be represented by < 1 , 0 >

Let the angle we wish to head be = [90 - 64] = 26 degrees

And let the magnitude of this direction be = 4

So.....using the Law of Cosines.....the length of the side connecting these two vectors is :

l^2 = 4^2 + 1 - 2(4)cos 26 = 9.809647629608

So.....l = 3.1320357005640916

And using the Law of Sines, the angle between the "flowing water" vector and this vector is given by :

sinΘ/4 = sin26/ 3.1320357005640916

sin-1 ( 4 *sin 26 / 3.1320357005640916) = Θ =145.954257197431°

But, because vectors will follow the law of parallelograms,, we need to proceed on an angular line that is supplementary to this = 34.045742802569° at a rate of 3.1320357005640916 mph

Let us prove that this is correct.....

The x component of the resultant is given by 1 + 3.1320357005640916*cos( 34.045742802569 ) = 3.5951761851965269745587204588

And the y component of the resultant is given by 3.1320357005640916*sin( 34.045742802569) = 1.7534845871557995584901543276

the magnitude of the resultant is given by √[ 3.5951761851965269745587204588^2 + 1.7534845871557995584901543276^2 ] = 4 which is correct

And the angle of the resultant is given by

tan^{-1} ( 1.7534845871557995584901543276/3.5951761851965269745587204588) = 26° ..which is also correct.....

And converting 34.045742802569° to a heading, we get N 55.954257197431 E at a speed of 3.1320357005640916 mph

Rounded, we have N 56° E at 3 mph........

Here's a pic of the actual situtation.........

CA is the resultant, CB is the " flowing water" vector, and CD is the line of travel we need to take to reach "A"

The "flowing water' vector serves to pull our "maintained" speed and direction towards the resultant vector...

CPhill
May 5, 2015

#2**+10 **

Best Answer

We can solve this with a "tip-to-tail" method.

Let the "flowing water " vector be represented by < 1 , 0 >

Let the angle we wish to head be = [90 - 64] = 26 degrees

And let the magnitude of this direction be = 4

So.....using the Law of Cosines.....the length of the side connecting these two vectors is :

l^2 = 4^2 + 1 - 2(4)cos 26 = 9.809647629608

So.....l = 3.1320357005640916

And using the Law of Sines, the angle between the "flowing water" vector and this vector is given by :

sinΘ/4 = sin26/ 3.1320357005640916

sin-1 ( 4 *sin 26 / 3.1320357005640916) = Θ =145.954257197431°

But, because vectors will follow the law of parallelograms,, we need to proceed on an angular line that is supplementary to this = 34.045742802569° at a rate of 3.1320357005640916 mph

Let us prove that this is correct.....

The x component of the resultant is given by 1 + 3.1320357005640916*cos( 34.045742802569 ) = 3.5951761851965269745587204588

And the y component of the resultant is given by 3.1320357005640916*sin( 34.045742802569) = 1.7534845871557995584901543276

the magnitude of the resultant is given by √[ 3.5951761851965269745587204588^2 + 1.7534845871557995584901543276^2 ] = 4 which is correct

And the angle of the resultant is given by

tan^{-1} ( 1.7534845871557995584901543276/3.5951761851965269745587204588) = 26° ..which is also correct.....

And converting 34.045742802569° to a heading, we get N 55.954257197431 E at a speed of 3.1320357005640916 mph

Rounded, we have N 56° E at 3 mph........

Here's a pic of the actual situtation.........

CA is the resultant, CB is the " flowing water" vector, and CD is the line of travel we need to take to reach "A"

The "flowing water' vector serves to pull our "maintained" speed and direction towards the resultant vector...

CPhill
May 5, 2015