# Problem involving displacement vectors

1. Jan 24, 2006

### frankfjf

Alright, here's the problem:

A car is driven east for a distance of 41 km, then north for 27 km, and then in a direction 28° east of north for 26 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

I'm not getting the right answer for part a, and I figure if part a is wrong, I'm going to have the same luck with part b.

From what I can tell, this problem is asking to obtain the total displacement by adding the vector components of each vector. I get the following numbers:

Vector 1: 41sin(0 degrees) = 0, 41cos(0 degrees) = 41

Vector 2: 27sin(90 degrees) = 27, 27cos(90 degreeS) = 0

Vector 3: 26sin(28) = 12.2, 26cos(28) = 23

Net X component: 41 + 0 + 23 = 64

Net Y component: 0 + 27 + 12.2 = 39.2

Then when I take the square root of 64^2 + 39.2^2 I get a wrong answer for the magnitude. What am I doing wrong?

2. Jan 24, 2006

### finchie_88

I think it's this bit that's wrong, it says 28 degrees east of north (see diagram below - this is my intepretation), so using a little trig, does that not make the x-component 26sin28 and the y component 26cos28, then carry on as normal, and you should get the right answer.
| / Where the vertical line is north, and the angle A is 28 degrees.
| /
|A /
| /
|/

Last edited by a moderator: Jan 24, 2006