Find the position vector of a particle whose acceleration vector is a = (1, 2) with an initial velocity vector (0, 0) and initial position vector (1, 0). (10 points)

a = <1,2>

v = <t+c1,2t+c2>
v(0) = <0,0> so c1=c2=0
v = <t,2t>

r = <1/2 t^2+c1, t^2+c2>
r(0) = <1,0> so c1=1, c2=0
t = <1/2 t^2+1, t^2>

is that the answer or do we have to plug in numbers from there?

To find the position vector of a particle given its acceleration vector, initial velocity vector, and initial position vector, we can use the kinematic equation:

x(t) = x0 + v0t + (1/2)at^2

where x(t) is the position vector at time t, x0 is the initial position vector, v0 is the initial velocity vector, a is the acceleration vector, and t is the time.

In this case, we are given the following values:
a = (1, 2),
v0 = (0, 0),
x0 = (1, 0).

Substituting the given values into the equation, we get:

x(t) = (1, 0) + (0, 0)t + (1/2)(1, 2)t^2

Since t = 0 gives us the initial position vector, we can ignore the (0, 0)t term.

So the position vector at any time t is:

x(t) = (1, 0) + (1/2)(1, 2)t^2

Simplifying the equation, we get:

x(t) = (1 + (1/2)t^2, (1/2)t^2)

Therefore, the position vector of the particle is (1 + (1/2)t^2, (1/2)t^2).