2D Kinematics I tried so hard

In summary, the car has a velocity of 17.3 m/s, while the truck has a velocity of 20.5 m/s which is directed 48.5 ° south of east. The resultant velocity is 14.496 m/s and the direction is 10.948 degrees.
  • #1
lmbiango
21
0
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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  • #2


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.
 
  • #3


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


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.
 
  • #5


you're right, thanks, I got it now!
 

Related to 2D Kinematics I tried so hard

1. What is 2D kinematics?

2D kinematics is the study of motion in two dimensions, taking into account both the direction and magnitude of an object's movement.

2. How is 2D kinematics different from 1D kinematics?

1D kinematics only considers motion in a straight line, while 2D kinematics takes into account motion in two different directions.

3. What are some common examples of 2D kinematics in everyday life?

Some common examples of 2D kinematics include throwing a ball, a car turning on a curved road, or a person walking in a zigzag pattern.

4. How is 2D kinematics used in scientific research?

2D kinematics is used in a variety of scientific fields, such as physics, engineering, and biomechanics, to study the movement of objects and analyze forces and accelerations in different directions.

5. What are some key equations used in 2D kinematics?

The most commonly used equations in 2D kinematics include the equations for displacement, velocity, and acceleration in two dimensions, as well as the Pythagorean theorem to calculate the magnitude of displacement or velocity.

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