Kinematics in 2 dimensions; position vector problem.

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Homework Help Overview

The discussion revolves around a kinematics problem involving a particle moving in a horizontal plane with constant acceleration. The original poster presents a scenario where the particle starts at the origin with an initial velocity and undergoes acceleration over a specified time, leading to questions about finding acceleration and the position vector at a later velocity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration and the position vector using different velocity values and time intervals. There is confusion regarding the correct time to use for the second part of the problem, as well as the implications of constant acceleration on the calculations.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions made about time and velocity. Some guidance has been offered regarding the use of previously calculated acceleration and the relationship between velocity and time. There is recognition of confusion among participants, but no explicit consensus has been reached.

Contextual Notes

There is a noted lack of clarity in the problem statement regarding the timing of the velocity change, which has led to misunderstandings. Participants are also grappling with the implications of constant acceleration on their calculations.

Purity
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Homework Statement



"A particle moves in the horizontal plane that contains the perpendicular unit vectors i and j. Initially it is at the origin and has velocity 18ims^-1. After accelerating for 10 seconds its velocity is (30i + 8j)ms^-1. Assume that the acceleration of the particle is constant.

a) Find the acceleration of the particle
b) Find the position vector of the particle when the velocity is (36i + 12j)ms^-1.

2. The attempt at a solution

For (a) i managed to find out the acceleration being (1.2i + 0.8j)ms^-2 using v=at+u and rearranging it to a=(v-u)/t.

But for (b) it says the answer is 405i + 90j

I've tried it over and over again but I don't think the question was specific enough (probably just me not understanding it).

I've tried it this way using:

u=18i + 0j
v=36i + 12j
t=10
R (position vector; i used R=(t/2)(u+v))=5(u+v) which gives the vector R= 240i + 40j

I've tried it by replacing the initial velocity with 36i + 12j but that didn't work either. I can probably work out the question if I understood what it basically is asking for.

Someone Please run me through this, I'm not good at mechanics :'(
 
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Purity said:
u=18i + 0j
v=36i + 12j
t=10
R (position vector; i used R=(t/2)(u+v))=5(u+v) which gives the vector R= 240i + 40j

You have the wrong value of t. At t = 10 the velocity vector is 30i + 8j m/s, not 36i + 12j m/s. What would t have to be in order for the velocity vector to be 36i + 12j m/s?
 
eumyang said:
You have the wrong value of t. At t = 10 the velocity vector is 30i + 8j m/s, not 36i + 12j m/s. What would t have to be in order for the velocity vector to be 36i + 12j m/s?

I only said t=10 because that was the only time value in the question and i couldn't use acceleration from my previous answer because I'm using a different velocity value now.

I'm so confused...
 
Purity said:

Homework Statement



"A particle moves in the horizontal plane that contains the perpendicular unit vectors i and j. Initially it is at the origin and has velocity 18ims^-1. After accelerating for 10 seconds its velocity is (30i + 8j)ms^-1. Assume that the acceleration of the particle is constant.

a) Find the acceleration of the particle
b) Find the position vector of the particle when the velocity is (36i + 12j)ms^-1.

2. The attempt at a solution

For (a) i managed to find out the acceleration being (1.2i + 0.8j)ms^-2 using v=at+u and rearranging it to a=(v-u)/t.

But for (b) it says the answer is 405i + 90j

I've tried it over and over again but I don't think the question was specific enough (probably just me not understanding it).

I've tried it this way using:

u=18i + 0j
v=36i + 12j
t=10
R (position vector; i used R=(t/2)(u+v))=5(u+v) which gives the vector R= 240i + 40j

I've tried it by replacing the initial velocity with 36i + 12j but that didn't work either. I can probably work out the question if I understood what it basically is asking for.

Someone Please run me through this, I'm not good at mechanics :'(

If \vec{v}_0 is the initial velocity vector and \vec{a} is the (constant) acceleration (both of which you know from part (a)), the velocity vector v(t) and position vector r(t) at time t, starting from the origin, are
\vec{v}(t) = \vec{v}_0 + \vec{a} t,\\<br /> \vec{r}(t) = \vec{v}_0 t + \frac{1}{2} \vec{a} t^2.<br />

RGV
 
It is a constant acceleration.
Use equation v2=u2+2as
 
Purity said:
I only said t=10 because that was the only time value in the question and i couldn't use acceleration from my previous answer because I'm using a different velocity value now.

I'm so confused...
Yes, you are. You can use acceleration from your previous answer- you have to- you are told that the acceleration is constant. Having a different velocity is due to the acceleration not being zero, not that the acceleration changes.
 
Oh i see :) I got confused because when i did pure I was taught that you couldn't divide matrices (i was looking at r=(v^2-u^2)/2a) and i was working all these out in matrix form (as a column vector) and it didn't occur to me to dividing it through because i was deliberately avoiding that. I tried dividing the i and j units and it came to the right answer.

Thank you :)
 
Shouldn't stumble into this problem again :D Thank you!
 

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