Solving Vector Problems in Motion: Finding Displacement and Velocity

  • Thread starter Thread starter integra2k20
  • Start date Start date
  • Tags Tags
    Vector
Click For Summary

Homework Help Overview

The discussion revolves around vector problems related to motion, specifically focusing on calculating displacement and velocity in two different scenarios involving a motorist's journey and a particle's motion under acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore methods for calculating total vector displacement and average velocity, with one participant expressing difficulty in understanding vector components. Another participant questions the correctness of their derived position vector based on given acceleration and initial velocity.

Discussion Status

Some participants have offered guidance on interpreting vector components and the importance of direction in calculations. There is an ongoing exploration of assumptions regarding the problem setup, particularly concerning the directions of motion and the potential for errors in the problem statement.

Contextual Notes

Participants note the absence of direction information in the problem statements, which is essential for accurately determining position vectors. There are mentions of potential errors in the homework software being used.

integra2k20
Messages
35
Reaction score
0
I am just aweful at this vector stuff, i can NEVER get it right.

A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.20 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.20 min trip, find the following values.

(a) total vector displacement

(b) average speed i got this part already, easy

(c) average velocity (include the magnitude)

total displacement i got 6291 meters (rounded). i did this by figuring out each x and y coordinate for the vectors, then adding them up, and finding the hypotenuse of the resulting vector. apparentlys its not right. As for the velocity part, i was unsure of how to go about starting it...
 
Physics news on Phys.org
also for this question:

A particle initially located at the origin has an acceleration of = 3.00 m/s2 and an initial velocity of 0 = 4.00 m/s.

(a) Find the vector position at any time t (where t is measured in seconds).

i put that the vector was (3.00t)i + (4.00t^2)j, apparently the j part is wrong and I am not sure why...
 
integra2k20 said:
also for this question:

A particle initially located at the origin has an acceleration of = 3.00 m/s2 and an initial velocity of 0 = 4.00 m/s.

(a) Find the vector position at any time t (where t is measured in seconds).

i put that the vector was (3.00t)i + (4.00t^2)j, apparently the j part is wrong and I am not sure why...

In general, the vector of position for a particle moving in the xy plane is given by [tex]\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}[/tex], where [tex]x(t)=x_{0}+v_{x0}t+\frac{1}{2}a_{x}t^2[/tex] and [tex]y(t)=y_{0}+v_{y0}t+\frac{1}{2}a_{y}t^2[/tex]. Now, you gave values of the acceleration and initial velocity, but you did not give any directions, which are needed to find the position vector.

Regarding the first post, question (a), I think the total displacement means the displacement with respect to the origin (i.e. the point from which the travel started), so it equals the resultant vector.
 
Last edited:
radou said:
In general, the vector of position for a particle moving in the xy plane is given by [tex]\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}[/tex], where [tex]x(t)=x_{0}+v_{x0}t+\frac{1}{2}a_{x}t^2[/tex] and [tex]y(t)=y_{0}+v_{y0}t+\frac{1}{2}a_{y}t^2[/tex]. Now, you gave values of the acceleration and initial velocity, but you did not give any directions, which are needed to find the position vector.

i copied/pasted the question exactly from my webassign, i guess it may be an error in the book? there's been a few errors before, but they really don't give you any directions associated w/ that problem.
 
Assuming that there is no error, you should be able to conclude from the context what the directions are. I would assume it is motion among a straight line, i.e. the direction is understood to have the direction of [tex]\vec{i}[/tex].
 
radou said:
Assuming that there is no error, you should be able to conclude from the context what the directions are. I would assume it is motion among a straight line, i.e. the direction is understood to have the direction of [tex]\vec{i}[/tex].

if that's the case, what am i doing wrong with the vector?
 
integra2k20 said:
if that's the case, what am i doing wrong with the vector?

If that's the case, you only have [tex]\vec{r}(t)=x(t)\vec{i}=(x_{0}+v_{0}t+\frac{1}{2}at^2)\vec{i}[/tex]. Plug in the initial speed, the acceleration, and the initial position (equals zero if the particle starts from the origin).
 
radou said:
If that's the case, you only have [tex]\vec{r}(t)=x(t)\vec{i}=(x_{0}+v_{0}t+\frac{1}{2}at^2)\vec{i}[/tex]. Plug in the initial speed, the acceleration, and the initial position (equals zero if the particle starts from the origin).

there's definitely an error in the software or something (i do my homework via webAssign)...it said that, for the i component of the vector, just 3.00 was right.

ill print that one out and ask my teacher tommorow, in the meantime, any suggestions on the first question i posted?
 
If you can figure out the lengths travelled, you know the directions and can sovle much of it graphically. The point being, for vector problems, always draw them to scale. Use graph paper, a ruler and a protractor.
 
Last edited:
  • #10
got it, thanks for the advice :)
 

Similar threads

Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
16K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
26
Views
4K
  • · Replies 2 ·
Replies
2
Views
993
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K