college mechanics

**divine_mail** · August 8th 2008, 10:53 AM

A particle of mass 2kg moves in the xy plane under the action of a force F where F = 2i-6j
initially, the particle is at the point whose position vector is i+j with a velocity vector i-j
find the position vector of the particle at any time t

Much Appreciated

**skeeter** · August 8th 2008, 12:41 PM

in the x-direction ...

$a_x = \frac{F_x}{m} = 1$

since the force remains constant ...

$x = x_0 + v_{0x} t + \frac{1}{2}a_x t^2$

$x = 1 + t + \frac{1}{2}t^2$

in the y-direction ...

$a_y = \frac{F_y}{m} = -3$

since the force remains constant ...

$y = y_0 + v_{0y} t + \frac{1}{2}a_y t^2$

$y = 1 - t - \frac{3}{2}t^2$

so ...

$\vec{r} = \left(1 + t + \frac{1}{2}t^2\right)\vec{i} + \left(1 - t - \frac{3}{2}t^2\right)\vec{j}$

**topsquark** · August 8th 2008, 07:33 PM

Originally Posted by divine_mail

A particle of mass 2kg moves in the xy plane under the action of a force F where F = 2i-6j
initially, the particle is at the point whose position vector is i+j with a velocity vector i-j
find the position vector of the particle at any time t

Much Appreciated

We can do this a bit more directly.
$\vec{a} = \frac{\vec{F}}{m} = \vec{i} - 3 \vec{j}$

So
$\vec{x} = \overrightarrow{x_0} + \overrightarrow{v_0}t + \frac{1}{2} \vec{a}t^2$

giving
$\vec{x} = (\vec{i} + \vec{j} ) + (\vec{i} - \vec{j} ) t + \frac{1}{2} ( \vec{i} - 3 \vec{j} ) t^2$
which is skeeter's answer.

-Dan

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