Calculating Displacement in Two-Dimensional Motion

  • Thread starter Thread starter Gamegoofs2
  • Start date Start date
  • Tags Tags
    Motion
AI Thread Summary
The discussion centers on calculating displacement in two-dimensional motion using vector components. The problem involves traveling at a speed of 1.2 m/sec for 45 minutes at a heading of 45 degrees, then changing direction to 330 degrees for 32 minutes. The user initially calculates the displacement for the first vector and arrives at 3240 meters, questioning its validity. Responses clarify that this value is reasonable given the time and speed, confirming that the calculation is correct. Overall, the conversation emphasizes understanding vector addition in two-dimensional motion.
Gamegoofs2
Messages
3
Reaction score
0
I'm having trouble with adding two-dimensional motion. Here's an example of the type of problem I'm having trouble with:

A map gives you the following directions:

Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees. Turn to a heading of 330 degrees and travel at the same speed for 32 minutes.
What will your displacement be if you follow these directions?

I was taught to use this equation:
\Deltax = initial velocity * time + 1/2 * acceleration * time^2
and
Ax= initial velocity * cos()
Ay= initial velocity * sin()
and the same for the the second vector.

I started out with trying to find the first vector or vector A as I called it.
\Deltax= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got \Deltax=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.
 
Physics news on Phys.org
Welcome to PF!

Hi Gamegoofs! Welcome to PF! :smile:

(have a delta: ∆ :wink:)
Gamegoofs2 said:
Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees.

I started out with trying to find the first vector or vector A as I called it.
\Deltax= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got \Deltax=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.

No, that's ok …

and it's not high, it's only 3km after nearly an hour! :wink:
 
Thanks so much!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top