# Vectors and Magnitudes

1. Sep 11, 2007

### chaotixmonjuish

A particle moves in the x-y plane with constant acceleration. At time t = 0 s, the position vector for the particle is d = 1.8 m x + 4.2 m y. The acceleration is given by the vector a = 1.5 m/s2 x + 6.2 m/s2 y. The velocity vector at time t = 0 s is v = 1.7 m/s x - 1.7 m/s y.

What is the magnitude of the position vector at time t = 7.5 s?

I tried setting up two equations:
F(x)=1.8+1.7x+1.5x^2
F(y)=4.2-1.7y+6.2x^2

This yielding something like 340. That answer didn't bode well with lon capa. I'm just not sure how to set up an equation to solve this.

2. Sep 11, 2007

### learningphysics

Just use the equation

d = x0 + v0*t + (1/2)at^2

Use that in the horizontal direction... then in the vertical direction.

3. Sep 11, 2007

### chaotixmonjuish

That's where my problem is. I'm not sure how to seperate x and y.

4. Sep 11, 2007

### learningphysics

I might be misunderstanding the problem... is this:

a = 1.5 m/s2 x + 6.2 m/s2 y

supposed to be:

$$\vec{a} = 1.5\vec{a_x} + 6.2\vec{a_y}$$ where $$\vec{a_x}$$ and $$\vec{a_y}$$ are unit vectors in the x and y directions?

I'm assuming they are just unit vectors... if they are supposed to be x and y, then acceleration isn't constant...

5. Sep 11, 2007

### chaotixmonjuish

I myself don't even know. This is the joys of lon capa.

6. Sep 12, 2007

### learningphysics

I'm guessing they are just unit vectors... in case you just divide the problem into two parts... first deal with horizontal displacement (ie x direction)

x0 = 1.8
v0 = 1.7
a = 1.5

hence x = 1.8 + 1.7t +(1/2)1.5t^2

calculate that for t=7.5... that gives x at 7.5

do the same thing in the y direction... that gives y at 7.5

Using the x and y you calculated... get the magnitude of the position vector using pythagorean theorem