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[milanbaros]

But just because its simple doesn't mean I can do it. I have difficulty understanding how initial position, velocity and acceleration can all have co-ordinates on the same graph. Here's the question.

In this question the unit of length is the metre (sorry americans) and the time is in seconds.

An object has initial position (2,-1) and initial velocity (-1,4).

It has a constant acceleration of (2,5)

i) Calculate the initial speed of the object.

ii) Calculate the object's velocity and position after 4 seconds.

- I had to write it as (2,-1) even though they should be written one above the other.

- Please may someone also explain to me how they can all coexist on the one graph.

Thanks

BTW, this isn't homework in the true sense, more revision.

 

[Papi]

The initial velocity is a vector from (2, -1) to (-1, 4).
The initial acceleration is a vector from (-1, 4) to (2, 5).
Or i'm completely wrong.

 

[Dilbert]

Don't confuse Cartesian coordinates and vector notation. (To be fair, the confusion is the fault of whatever dumbass came up with this problem...but oh well.)

When you say that the initial position of the object is (2, -1), you mean that you can draw that object as a point on a Cartesian plane with x = 2 and y = -1. However, when you say that the velocity is (-1, 4), you mean that the position of the object will shift to the left (x-direction) by 1 unit per unit time and shift upwards (y-direction) by 4 units per unit time.

In other words, if the velocity was constant, the position of the particle at any time t would be:

x = 2 - t
y = -1 + 4t

In this case, though, you have a nonzero acceleration which is expressed -- you guessed it -- in vector notation.

My recommendation is to get rid of the confusing notation and parametrize the problem the way I've started to do, with separate equations for x(t) and y(t). You can integrate from the components of velocity to get the equations for v(t) and a(t) in each dimension, and then calculate the answer.

Good luck...

 

[Dilbert]

The initial velocity is a vector from (2, -1) to (-1, 4).
The initial acceleration is a vector from (-1, 4) to (2, 5).
Or i'm completely wrong.

Completely wrong. Sorry.

 

[slayn]

going by the way the problem is written, I would think parameterising (I hate that word... so hard for me to think how to spell or even say it) it and approacing it form that would be 'too difficult' at this point in time.

At least by this lone question, I would guess all he was supposed to do was find the speed by sqrt(1^2+4^2)

and then finding velocity and position by just adding values in steps. like 1 second later
position is (2,-1) '+' (-1,4) = (1,3)
accel is (-1,4) '+' (2,5) = (1,9)

and so on until you get to 4 seconds.

I know it not technically right... but its the kind of wrong answer I would think is expected for the 'correct' answer for low level stuff. But I could be wrong.


as for how they can all be on the same graph... I wouldn't try putting acceleration on there. Like you could do position and velocity by putting a dot at (2,-1) and then velocity by drawing a vector from (2,-1) to (1,3)

accel I guess I would raw as some kond of force. Use another color of pencil and draw 2,5 vectors everywhere or something.

 

[Dilbert]

But I could be wrong.

Yes, you are.

I don't mean to be rude, but did you see what I wrote? The ( , ) notation means a vector. Even though we think of (x, y) as being a "point" in Cartesian space, it is more properly thought of as the value of the displacement vector with respect to the Cartesian origin.

 

[slayn]

and where did I not treat velcoity/accel as a vector?

a velcoity vector of -1,4 would move it 'left' 1 and 'up' 4 after one second. Hence after one second the new position would be 1,3 ignoring acceleration effects for that one second. Which is what I seem to recall teachers having us do when we were very first introduced to this stuff. IE. learning a wrong answer because its simpler and then eventually showing why its wrong. And I was taking a guess that he was at this level.

edit:

I mean if it were ME I would break x/y portions apart and do
d = vi*t + (1/2)at^2
vf^2 = vi^2 + 2ad

but I made a guess that thats not how the teacher wanted them to approach this problem.