(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

R = sqrt(A^2 + B^2)

Ax = cosA, Ay = sinA

tan = y/x

3. 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.

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# Magnitude and direction of resultant

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