# Thread: Need help in physics vector problem

1. ## Need help in physics vector problem

A car is driven east for a distance of 41 km, then north for 25 km, and then in a direction 30° east of north for 23 km. Determine (a) the magnitude (in km) 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.

The answer I got is 71km and the angle I got is 30.93 degrees north east. However the computer says that is wrong.

To get x and y, I added up all the x and y together.
x = 41 + 0 + 23cos30
y = 0 + 25 + 23sin30

This should yield x = 60.91 and y = 36.5 right?

To get answer for part a, you would square 60.91 and 36.5 and then take the square root of them. So that comes out to be 71km.

Well for part b, I took the inverse tan of 36.5/60.91.

Please tell me what I did wrong! The computer is driving me nuts telling me I did it wrong.

2. Originally Posted by florx
A car is driven east for a distance of 41 km, then north for 25 km, and then in a direction 30° east of north for 23 km. Determine (a) the magnitude (in km) 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.

The answer I got is 71km and the angle I got is 30.93 degrees north east. However the computer says that is wrong.

To get x and y, I added up all the x and y together.
x = 41 + 0 + 23cos30
y = 0 + 25 + 23sin30

This should yield x = 60.91 and y = 36.5 right?

To get answer for part a, you would square 60.91 and 36.5 and then take the square root of them. So that comes out to be 71km.

Well for part b, I took the inverse tan of 36.5/60.91.

Please tell me what I did wrong! The computer is driving me nuts telling me I did it wrong.
30 degrees east of north is 60 degrees relative to the + x-axis