# Homework Help: Simple Coordinate Problem

1. Sep 25, 2009

### RoganSarine

"Simple" Coordinate Problem

1. The problem statement, all variables and given/known data
The following happens on a 2D X-Y Plane.

A particle accelerates at {3t m/s2}i + 4t m/s2}j where t = seconds

At t = 0, the position of the particle is {20.0 m}i + {40.0m}j

At t = 0, the velocity of the particle is {5.00 m/s}i + {2.00 m/s}j

At t = 4,

(a) What is the position vector in unit-vector notation
(b) What is the angle between its direction of travel and the positive direction of the x-axis

2. Relevant equations
dx = Vox*t + .5(a)t^2
Dy = Voy*1 + .5(a)t^2

3. The attempt at a solution

dx = Vox*t + .5(a)t^2
dx = 5(4) + .5(3(4)) (4)^2
dx = 20 + 6(16)
dx = 116

Dy = Voy*1 + .5(a)t^2

Dy = 2(4) + .5(4(4)) (4)^2
Dy = 8 + 8(16)
Dy = 136

Therefore, it's position vector should be

P = {136 m}i + {176 m}j

If we remember to add on the initial displacement.

The answer is {72.0 m)i + {90.7 m}j

2. Sep 26, 2009

### D H

Staff Emeritus
Re: "Simple" Coordinate Problem

The relevant equations you cited are for constant acceleration. You do not have a constant acceleration here.

3. Sep 26, 2009

### RoganSarine

Re: "Simple" Coordinate Problem

Huh... Right. True. However, I don't ever remembering having to deal with anything that isn't constant acceleration.

4. Sep 26, 2009

### D H

Staff Emeritus
Re: "Simple" Coordinate Problem

First you should do is check that you have the problem correct. If it is correct as written, I suspect that you have learned something about integration.

5. Sep 26, 2009

### RoganSarine

Re: "Simple" Coordinate Problem

Yes, the problem is correct.

We have actually learned nothing about integration specifically, but I do know the basic calculus for integration and have a basic understanding of it.

But I do not see how it applies.

Basically, this question was not even assigned to us I don't think, I am simply going through all the questions my text gives me in the section, so... If you could give me a giant push or something, I'd appreciate it.

6. Sep 26, 2009

### D H

Staff Emeritus
Re: "Simple" Coordinate Problem

You know the acceleration as a function of time and the initial velocity. Acceleration is the derivative of velocity, so integrating acceleration with respect to time will yield velocity as a function of time. Integrating velocity will yield position.

7. Sep 26, 2009

### RoganSarine

Re: "Simple" Coordinate Problem

So in theory, wouldn't

ax(t) = 3t
vx(t) = (3/2)t^2 + c
px(t) = (3/6)t^3 + ct + d

If vx(0) = 5

then

vx(0) = (3/2)(0)^2 + c
5 = 0 + c
c = 5
vx(t) = (3/2)t^2 + 5

px(t) = (3/6)t^3 + ct + d
px(t) = (3/6)t^3 + 5t + d
px(0) = 20
px(0) = (3/6)(0)^3 + 5(0) + d
20 = 0 + 0 + d
d = 20

Therefore:

p(x) = (3/6)t^3 + 5t + 20
p(4) = (3/6)(64) + 20 + 20
p(4) = 32 + 40 = {72 m}i

Which... is the answer for the x coordinate. Thanks for the hint -- I knew it couldn't have been too hard once again just tired.