2D Kinematics I tried so hard

AI Thread Summary
A user is struggling to calculate the truck's velocity relative to the ground, given its velocity relative to a car and the car's velocity relative to the ground. Initial calculations using incorrect angles led to wrong results for both magnitude and direction. A suggestion was made to ensure consistent values in calculations, particularly using the correct angle of 48.5 degrees instead of 45 degrees. The user acknowledged the mistake and noted the need to recalculate the y-component based on the correct angle. The discussion emphasizes the importance of accuracy in vector components for solving 2D kinematics problems.
lmbiango
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2D Kinematics Please Help! I tried so hard

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 don't know what to do!
 
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Hi lmbiango,

lmbiango said:
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)

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


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.
 


lmbiango said:
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.


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.
 


you're right, thanks, I got it now!
 
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