Question

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of south.

I know the formulas but having trouble figuring out the degree of angle.

Ax=4(cos30)= 3.4641
Ay=4(sin30)= 2

Bx=3.7(cos20)= 3.477
By=3.7 (sin20)= 1.2654

Rx = 3.4641 + 3.477 = 6.9411
Ry = 2+1.2654 = 3.2654

R=7.6708

tan theta = Rx/Ry
= 6.9411/3.2654
=2.125650762540577
=0.037116634989 SW

Please let me know if I am doing this correct

Answer 1

R=7.2

49 degrees south of west

Answer 2

Srry bro ur on your own

Answer 3

I know this is dead, but to find the degree of an angle after you solved for Tan(theta), you just do Tan^-1(2.125650762540577) which would be 64.80563. That is the degree. Hope this helps that one random person who sees this one day.

Answer 4

Thanks Jacob's awesome!

Answer 5

[AI-generated answer — Explain Bot]

You are on the right track in using the correct formulas to find the x and y components of the vectors. However, there are some mistakes in the calculations.

To find the x and y components of the first vector:

Ax = 4.0 * cos(30°) = 3.4641 m/s (round to four decimal places)
Ay = 4.0 * sin(30°) = 2.0 m/s

To find the x and y components of the second vector:

Bx = 3.7 * cos(20°) = 3.4666 m/s (round to four decimal places)
By = 3.7 * sin(20°) = 1.2679 m/s (round to four decimal places)

Now, adding the x and y components:

Rx = 3.4641 + 3.4666 = 6.9307 m/s (round to four decimal places)
Ry = 2.0 + 1.2679 = 3.2679 m/s (round to four decimal places)

To find the magnitude of the resultant vector R:

R = sqrt(Rx^2 + Ry^2)
= sqrt((6.9307)^2 + (3.2679)^2)
= sqrt(47.99744 + 10.67268)
= sqrt(58.67012)
= 7.6667 m/s (round to four decimal places)

Now, to find the angle of the resultant vector:

tanθ = Ry / Rx
θ = tan^(-1)(Ry / Rx)
= tan^(-1)(3.2679 / 6.9307)
= tan^(-1)(0.4710)
= 25.98° (rounded to two decimal places)

Therefore, the angle of the resultant vector is approximately 26°.