The problem involves calculating the displacement of a car that travels a series of distances in different directions: 8 km east, 8 km north, and 16 km west. The original poster attempts to determine the resultant displacement using the Pythagorean theorem but expresses uncertainty about the completeness of their answer.
The discussion is ongoing, with participants providing insights on the distinction between average speed and average velocity. There is an exploration of the angle of the resultant vector, with some participants questioning the need for precise directional measurements. Guidance has been offered regarding the verification of angles and the importance of understanding vector components.
There is mention of homework constraints and the need for participants to avoid providing complete solutions. The original poster's confusion about the relationship between distance traveled and displacement is also noted.
Medgirl314 said:Thanks for your replies!
Agreed, I'm never satisified until I find the answer. I drew a diagram, and I'm attempting to find the answer using the parallelogram method. However, my visualization skills don't seem to be kicking in. I have a diagram that does look like a parallelogram, and it can be divided into two right triangles. Working from either side, it appears that the resultant is 11.3, but that doesn't seem complete.
There's an interesting distinction between average speed and average velocity when the path is not a straight line. For average speed you take the total distance traveled along the path, but for average velocity you just look at the vector connecting start and end points - the path is irrelevant.Medgirl314 said:So the distance traveled would be 32 km and if the trip took three hours, the average veocity would be 10.6 km/h NE, correct? And the speed would be 10.6 km/h?
Medgirl314 said:Are you saying that my answer is correct, but the NE part isn't needed? My physics teacher noted that it was important, though that may go against the norm.
Medgirl314 said:Right, so would that be ENE? Judging by the direction of my resultant?
Medgirl314 said:Right, so would that be ENE? Judging by the direction of my resultant?
Medgirl314 said:Okay, I see it now. I looked at my drawing, which was flawed. It would be NW. Or maybe phrashing it like my teacher does, WNW.
Medgirl314 said:Oh, I didn't know that! Thank you. My teacher just said "Well, this resultant is pointing slightly more to the right than up, so it's ENE." Which made sense, it just wasn't explained. The degree explanation helps. Mine was pointing NW because my drawng was flawed.
Medgirl314 said:Oops, I meant mine was pointing northeast, not northwest. It would be exactly NE, 45 degrees, the way I drew it. But that was the wrong drawing.
You really should verify it algebraically. What is the net displacement in the N direction? In the W direction? How do you deduce the angle from those two?Medgirl314 said:Oh! I thought we already confirmed that, so I wasn't sure what you were trying to show me. In that case, it would be a 45 degree angle, which I could verify with a protractor
Measuring the angle in the standard navigational way, degrees clockwise from N, you would use tan(θ) = eastern displacement/northern displacement.Medgirl314 said:It would be exactly between north and west, so -45 degrees. Is there a specific formula I'm suppossed to be using? The answer, degree-wise, seems a bit obvious to me because the angle is a 45 degree angle, and it is pointing in the negative direction. Thanks again!
cepheid said:FaraDazed, we don't give complete solutions or give away answers on PF. Since you've done so, we have no idea if the OP understands how to determine the direction.
Medgirl314 said:Ah, okay. Thank you! My physics teacher hasn't gone into that yet, and so would perfer that I give the direction according to N/S and so on, but that's a great aid!