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

[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.

 

[learningphysics]

Just use the equation

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

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

 

[chaotixmonjuish]

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

 

[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...

 

[chaotixmonjuish]

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

 

[learningphysics]

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

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