# 2D Kinematics I tried so hard

1. Sep 11, 2008

### lmbiango

Relative to the ground, a car has a velocity of 17.3 m/s, directed due north. Relative to this car, a truck has a velocity of 20.5 m/s directed 48.5 ° south of east. Find the (a) magnitude and (b) direction of the truck's velocity relative to the ground. Give the directional angle relative to due east.

I got:

Vcg = 0x + 17.3y
Vtg = 20.5 cos (45) - 20.5 sin(45)

The resultant is: 14.496x + 2.804y
So, by the Pythagorean theorem, Vtg = 14.765

(But that's wrong)

And the direction would be: Inverse tangent of (2.804/14.496)

so it's 10.948 degrees

And that's wrong too.

I also tried it with adding the 20.5 sin(45) on the Vtg part, but those answers are wrong too. I dont know what to do!

2. Sep 12, 2008

### alphysicist

Hi lmbiango,

I don't see why are you using 45 degrees here; they give a different angle in the problem.

3. Sep 12, 2008

### lmbiango

you're right, that was my first dumb mistake... but even after getting the correct number for Vtg, which I now have 13.584, the angle is still incorrect. If I do, inverse tangent of (2.804/13.723) = 11.548, that is wrong for the angle and I only have one try left.

4. Sep 12, 2008

### alphysicist

Looking at your numbers, I think you might be mixing up a few numbers.

You say the answer for Vtg was 13.584, but in your inverse tangent equation you have 13.723 in the denominator. These should be the same number and I think the 13.723 is correct, so you might want to check why you have different numbers.

In the numerator of your inverse tangent you use 2.804, which was the y component you found when using the incorrect angle of 45 degrees. You need the new total y component from using 48.5 degrees.

5. Sep 14, 2008