
#1
Sep1711, 07:52 AM

P: 4

A particle moves in the xy plane with constant
acceleration. At time zero, the particle is at x = 5.5 m, y = 2.5 m, and has velocity ṽ=(2.5 m/s) î+ (−2 m/s)ĵ . The acceleration is given by ā =(3 m/s2) î+ (7.5 m/s2)ĵ. What is the x component of velocity after 2.5 s? Answer in units of m/s What is the y component of velocity after 2.5 s? Answer in units of m/s What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 2.5 s? Answer in units of m The use of velocity and accelration in terms of vector units( i,j coordinates) instead of a single numerical measurement has me confused, especially since it is not in my text book. Avy suggestions? 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Sep1711, 08:57 AM

P: 3,627

Welcome to the board!!
I think you are forgetting this: The forum rules says that the questioner need to show his/her attempts before asking for any help. 



#3
Sep1711, 11:02 AM

P: 4

thanks for the response, here is kinda what I have done so far
ahh...attempts! I have tried to solve this several ways with multiple answers and still i get it wrong. 1st I tried to solve for v;v=at, using 3m/s2i as accel>then used v/time to get distance and add to x to get , wrong. I tried plotting the slopes of velocity and acceleration in conjunction with particle origin...got nowhere. I then tried a variation of my first approach using 2.5m/si as V=Δx/ΔT and solved for x, then added it to 5.5m, nope not that. Also have tried using Δx=Voî * t + 1/2aî*t^2 then added this to x=5.5m. Ultimately I just don't know what to do with the two Velocities and two Accelerations, I have come to learn that a vector can be described by a non scalar coordinate pair, not x and y, but î and ĵ which are I think are vector untis. I have only one problem that uses vector units rather than just a=3m/s^2, and because of v and a being given in terms of (#î)+(#ĵ) I don't know what to with these vector units. It wasn't discussed in class and the use of î and ĵ is not in my book, after only after some time on the internet did I figure out what the î and ĵ were. But I haven't found anything example that includes vector units for velocity and acceleration that are used with time. What I have now is basic; the answer is in m/s so it a velocity, time is given as a variable which is in seconds which when multiplied by accel in m/s^2 gives a velocity; and then I am confused as to what to do with because of there are two components of both v and a, although it would be nice if only the î parts were used since the answer is asking for the x component of of velocity(which for projectile motion is constant because there is no horizontal component of acceleration) This describes my attempts to solve this problem, and the frustration of not what to do with v and a when in vector units (since my textbook does not include this) when the actual problem is not that difficult. Any thoughts or suggestions would be greatly appreciated. 



#4
Sep1711, 12:49 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

vector units;velocity,accel, and time
Yes, i and j are unit vectors. The vector i has a magnitude of 1 and points in the positive xdirection. The vector j has a magnitude of 1 and points in the positive ydirection. Any vector in the xdirection can be expressed as a scaled version of i. Any vector in the ydirection can be expressed as a scaled version of j. Any vector in the plane can be expressed as a vector sum of x and ycomponents, each of which can be expressed in terms of unit vectors as described above. Because of this property, the vectors i and j are said to "span" the vector space and are known as "basis vectors" for that space.
You can think of this problem as two independent 1D problems. You have an acceleration for the xdirection, and an acceleration for the ydirection. You have an initial velocity for the xdirection, and an initial velocity for the ydirection. Compute the components separately at any time t, add them together (as a vector sum), and the resultant still gives you the total velocity vector at time t. 



#5
Sep1711, 12:54 PM

P: 3,627

(I am representing vectors in boldface.)
Okay, so you don't know much about vectors. I will give a brief of what is i and j. They are nothing more than unit vectors in the direction of x and y axis respectively. For z axis, the unit vector is k, but we don't have to deal with that right now. If you multiply a scalar quantity with them, for example [itex]5 x i[/itex], it becomes a vector of 5 magnitude in the xaxis direction since i is in the direction of xaxis. So our vectors is 5i. For example, you have a vectors which has both a x and y component, like: 7i+9j Then you can take them as independent vectors in their respective directions like 7i goes in the xaxis direction and the other goes in yxis direction. Now lets get back to the question. You have your velocity vector as (2.5 m/s)i+ (−2 m/s)j. As i explained above, you can take them in their respective directions. 2.5 m/s in xaxis direction and 2m/s in the yaxis direction. Similarly, you can do this for the acceleration vector. Now you have all the data for the motion in xaxis and yaxis direction, you can now apply the equations of motions. :) Your first approach is wrong because you took initial velocity as zero which is wrong. Here initial velocity in the direction of xaxis is 2.5m/s. Read the question properly "At time zero, the particle is at x = 5.5 m, y = 2.5 m, and has velocity ṽ=(2.5 m/s) î+ (−2 m/s)ĵ ." 



#6
Sep1711, 01:23 PM

P: 4

Thank you! you are right I don't know much about vectors. That is so simple, I thought that somehow the x and y axis information might have to be combined but then if I only used one is seemed like I would leave out half of the velocity; but using the information of one axis independent of the other makes sense since finding the x component of velocity only relates to the x axis. Now when I read the question the velocity information makes sense along with the initial velocity being 2.5 m/s. Thanks again, it seems so simple now.



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