2D Kinematics I tried so hard

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Homework Help Overview

The discussion revolves around a 2D kinematics problem involving the velocities of a car and a truck. The original poster seeks to determine the magnitude and direction of the truck's velocity relative to the ground, given specific directional angles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the components of velocity and the angles involved, with some questioning the use of an incorrect angle in calculations. There are attempts to clarify the correct values for the truck's velocity and its directional angle.

Discussion Status

Participants are actively engaging with the problem, identifying mistakes in calculations and discussing the implications of using the wrong angle. Some guidance has been provided regarding the need to ensure consistency in the values used for calculations.

Contextual Notes

There is a mention of limited attempts remaining for the problem, which adds pressure to resolve the calculations accurately. The original poster expresses concern over potentially incorrect answers and the need to verify their approach.

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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