How Can I Improve My Approach to Solving 2D Motion Problems?

  • Thread starter Thread starter Mono182
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
The discussion focuses on strategies for solving 2D motion problems, particularly through sketching and vector analysis. A user struggles with calculating the initial velocity of a ball thrown from a hot-air balloon and the average acceleration of a truck on a winding highway. Participants point out errors in the user's use of sine and cosine functions in their calculations, emphasizing the importance of correctly interpreting angles and converting units from km/h to m/s. The conversation highlights the need for accurate vector diagrams to visualize motion directions. Overall, the thread provides insights into common pitfalls in 2D motion problems and encourages careful attention to detail in calculations.
Mono182
Messages
11
Reaction score
0
My main problem right now is that i can't figure out most of the 2 dimensional problems. I try sketching it out but I'm finding it really hard. Can anyone help me figure out strategies to help me to beable to draw or sketch them out in the future.

1) A passenger in a hot-air balloon throws a ball with an intial unknown velocity. The ball accelerates at 9.8 m/s² [down] for 2.0 s, at which time its instantaneous velocity is 24 m/s [45º below the horizontal]. Determine the ball's initial velocity.

2) At 3:00 PM a truck traveling on a winding highway has a velocity of 82.0 km/h [38.2º E of N]; at 3:15 PM it has a velocity of 82.0 kn/h [12.7º S of E]. Assuming that +x is east and +y is north, determine the x- and y-components of the average acceleration during this time interval.
 
Physics news on Phys.org
Hi, sorry for dig up a old post. But i am stuck on question 2 (as above) as well.

This is what i did:

For x:

a= (v2 - v1)/ Δt
a= (82sin12.7 - 82cos38.2)/ 0.25
= -185.65 km/h(2)

For y

a=(v2-v1)/Δt
=(-82cos12.7 - 82sin38.2)/0.25
= -522.81 km/h(2)

the correct answer is for x: 0.009 m/s/s
y: -0.025 m/s/s

thank you.
 
Looks like you have sin and cos switched around.
 
can you speicify which part is switched incorrectly? It doesn't seem to be according to my vector diagram.
 
Everywhere that has either sin or cos.

p.s. welcome to PF :smile:
 
Thanks : )
but no matter how much i switch the sin or the cosine i still don't get the required answer.
 
Are you watching the units carefully, for both distance and time?
 
thanks for the replies be the way.

initiall the distance and time is in Km and Hours. So i converted the 15min into 0.25 hours and i plugged everything in. At the end i reconverted km/h to m/s to see if the answers match but they didn't. Would there be something wrong with my vector diagram? I have V1 pointing north east (assume north and east are both +) and i have v2 pointing in the south east direction (where Y= - and X= +)
 
Ste_0 said:
thanks for the replies be the way.
You're welcome.
At the end i reconverted km/h to m/s
Shouldn't that be km/h2 to m/s2?

Would there be something wrong with my vector diagram? I have V1 pointing north east (assume north and east are both +)
Yes, and it should be 38.2º from due NORTH.

... and i have v2 pointing in the south east direction (where Y= - and X= +)
Yes, and hopefully you have it 12.7º from due EAST.
 
Back
Top