# Magnitude of displacement vector

• May 2nd 2009, 10:56 AM
Solid8Snake
Magnitude of displacement vector
A jet-ski leaves the dock and travels 2km north. It then goes 1.4 km southeast and stops near a large boulder. What is the magnitude of the vector representing the jet ski's displacement from the dock to the boulder.

I did the question and my answer was 1.414284887km however the answer booklet says 1.1km, im I doing something wrong?

If yes plz guide me through this question.
• May 2nd 2009, 11:13 AM
VonNemo19
I would use the law of cosines for this onehttp://wpcontent.answers.com/math/d/...22fc14aede.png

So the dude goes 2km north and then makes a 135 degree turn to the right and goes 1.4km

we have everything we need.

cos135=2^2+1.4^2-a^2)/2[(2)(1.4)]=(4+1.96-a^2)/5.6

=5.96-a^2/5.6

Can you finish up?
• May 2nd 2009, 11:23 AM
Solid8Snake
well using the formula you gave i get 3.149571078, however the final answer is suppose to be 1.1km
• May 2nd 2009, 11:50 AM
VonNemo19
Try again, but this time let b=1.4 , c=2
• May 2nd 2009, 11:51 AM
aidan
Quote:

Originally Posted by Solid8Snake
A jet-ski leaves the dock and travels 2km north. It then goes 1.4 km southeast and stops near a large boulder. What is the magnitude of the vector representing the jet ski's displacement from the dock to the boulder.

I did the question and my answer was 1.414284887km however the answer booklet says 1.1km, im I doing something wrong?

If yes plz guide me through this question.

The angle from the dock to the point 2 km north to the boulder is 45 degrees.
The length of the hypotenuse of a triangle with sides of length 1 is $\displaystyle \sqrt 2$ or approx 1.4

The boulder point is 45 degrees east of north at a distance of about 1.4 km. or 1.010050506 km north and 0.989949494 km east of the dock.