Magnitude and direction of resultant

[jwxie]

Homework Statement

A car travels 20 km due north and then 35 km in a direction 60 degree west of north. Find the magnitude and direction of the car's resultant displacement.

Homework Equations

R = sqrt(A^2 + B^2)
Ax = cosA, Ay = sinA
tan = y/x

The Attempt at a Solution

This question was introduced and solved in the textbook before the author introduced Components of a Vector and Unit Vectors (i, j, z; Ax, Ay, Az).

I came back to this question as an attempt. The author used law of cosines in his deomonstration, as well as sin beta to solve the angle (direction).
He gave R = 48.2km at 38.9

I tried this problem with component method.
I drew a picture and I started the problem by listing components:

let the north 20km = A, 35km = B and their resultant = R

Ax = 0 (by all means, cos90 * 20 gives zero anyway)
Ay = sin90 * 20 = 20
Bx = cos60 * 35 = 17.5
By = sin60 * 35 = 30.3

** --> = vector
--> R = (Ax + Bx)i + (Ay + By)j
--> R x = Ax + Bx (17.5)
--> Ry = Ay + By (50.3) and
--> R = sqrt (Rx^2 + Ry^2)

In the end, R I got ~ 53 km, and for the degree, I got tan (Ry/Rx) = 70.7, but since the picture shows the direction is beyond 90, I say 180 - 70.7 = 109.3

Where did I do wrong?

Thanks.

 

[hage567]

Homework Statement

A car travels 20 km due north and then 35 km in a direction 60 degree west of north. Find the magnitude and direction of the car's resultant displacement.

Homework Equations

R = sqrt(A^2 + B^2)
Ax = cosA, Ay = sinA
tan = y/x

The Attempt at a Solution

This question was introduced and solved in the textbook before the author introduced Components of a Vector and Unit Vectors (i, j, z; Ax, Ay, Az).

I came back to this question as an attempt. The author used law of cosines in his deomonstration, as well as sin beta to solve the angle (direction).
He gave R = 48.2km at 38.9

I tried this problem with component method.
I drew a picture and I started the problem by listing components:

let the north 20km = A, 35km = B and their resultant = R

Ax = 0 (by all means, cos90 * 20 gives zero anyway)
Ay = sin90 * 20 = 20
Bx = cos60 * 35 = 17.5
By = sin60 * 35 = 30.3

** --> = vector
--> R = (Ax + Bx)i + (Ay + By)j
--> R x = Ax + Bx (17.5)
--> Ry = Ay + By (50.3) and
--> R = sqrt (Rx^2 + Ry^2)

In the end, R I got ~ 53 km, and for the degree, I got tan (Ry/Rx) = 70.7, but since the picture shows the direction is beyond 90, I say 180 - 70.7 = 109.3

Where did I do wrong?

Thanks.

I think your components for your B vector are wrong. Your Bx component should be 35sin60, not 35cos60. Likewise, your By component should be 35cos60 instead of 35sin60. Draw it out carefully and re-check your trig.

 

[kerimek]

Your approach using component method is correct. However, there are a few errors in your calculations.

Firstly, the magnitude of the resultant should be R = sqrt (Rx^2 + Ry^2) = sqrt ((17.5)^2 + (50.3)^2) = 53.7 km. This is slightly different from the value you obtained.

Secondly, to find the direction of the resultant, you need to use the inverse tangent function (arctan) to find the angle, not just the tangent function. So it should be tan^-1(Ry/Rx) = tan^-1(50.3/17.5) = 70.8 degrees.

Lastly, to find the direction of the resultant, you need to add the angle of the second vector (35 km in a direction 60 degrees west of north) to the angle you just calculated (70.8 degrees). So the final angle would be 70.8 degrees + 60 degrees = 130.8 degrees.

Therefore, the magnitude and direction of the car's resultant displacement would be 53.7 km at 130.8 degrees west of north.

I hope this helps clarify your doubts. Keep up the good work!