Kinematics in 2 dimensions; position vector problem.

In summary, the particle moves with a constant acceleration for 10 seconds and then has a position vector of (240i + 40j).
  • #1
Purity
7
0

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 :'(
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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...
 
  • #4
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 [itex]\vec{v}_0[/itex] is the initial velocity vector and [itex] \vec{a}[/itex] 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
[tex] \vec{v}(t) = \vec{v}_0 + \vec{a} t,\\
\vec{r}(t) = \vec{v}_0 t + \frac{1}{2} \vec{a} t^2.
[/tex]

RGV
 
  • #5
It is a constant acceleration.
Use equation v2=u2+2as
 
  • #6
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.
 
  • #7
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 :)
 
  • #8
Shouldn't stumble into this problem again :D Thank you!
 

What is kinematics in 2 dimensions?

Kinematics in 2 dimensions is a branch of physics that studies the motion of objects in a two-dimensional space. It involves analyzing the position, velocity, and acceleration of objects as they move in two perpendicular directions.

What is a position vector?

A position vector is a mathematical representation of an object's location in a two-dimensional space. It is a directed line segment that starts at the origin and ends at the object's position, with its length and direction representing the distance and direction from the origin to the object.

How do you solve problems involving position vectors?

To solve problems involving position vectors, you first need to identify the initial and final positions of the object and the direction and magnitude of its motion. Then, use mathematical equations and vector operations to calculate the object's displacement, velocity, and acceleration.

What is the difference between displacement and distance?

Displacement is a vector quantity that refers to the change in an object's position from its initial to its final position. On the other hand, distance is a scalar quantity that refers to the total length of the path traveled by the object, regardless of its direction.

Can kinematics in 2 dimensions be applied to real-world situations?

Yes, kinematics in 2 dimensions can be applied to real-world situations, such as projectile motion, motion of vehicles, and motion of objects on an incline. It is a fundamental concept in physics and is used in various fields, including engineering, sports, and astronomy.

Similar threads

  • Introductory Physics Homework Help
Replies
30
Views
490
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
891
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Replies
14
Views
1K
Replies
12
Views
926
  • Introductory Physics Homework Help
Replies
16
Views
2K
Back
Top