The vector position of a 3.45 g particle moving in the xy plane varies in time according to

1 = 3 + 3t + 2t2
where t is in seconds and is in centimeters. At the same time, the vector position of a 5.00 g particle varies as 2 = 3 − 2t2 − 6t.

Determine the vector position of the center of mass at t = 2.80.

your two equations do not reflect any positional information, as x= or y=

I have no idea what your equations mean. Vectors require directions.

To determine the vector position of the center of mass, we need to find the average position of the two particles at a specific time.

Let's start by finding the position vectors of the two particles at t = 2.80.

For the first particle:
r₁ = 3 + 3t + 2t²
= 3 + 3(2.80) + 2(2.80)²

We can plug in t = 2.80 into the equation to find r₁.

r₁ = 3 + 3(2.80) + 2(2.80)²
= 3 + 8.40 + 15.68
= 27.08 cm

Therefore, the position vector of the first particle at t = 2.80 is 27.08 cm.

For the second particle:
r₂ = 3 - 2t² - 6t
= 3 - 2(2.80)² - 6(2.80)

We can also plug in t = 2.80 into the equation to find r₂.

r₂ = 3 - 2(2.80)² - 6(2.80)
= 3 - 2(7.84) - 16.80
= 3 - 15.68 - 16.80
= -29.48 cm

Therefore, the position vector of the second particle at t = 2.80 is -29.48 cm.

Now, we can find the position vector of the center of mass by taking the average of the position vectors of the two particles.

r_cm = (r₁ + r₂) / 2
= (27.08 + (-29.48)) / 2
= -1.40 / 2
= -0.70 cm

Therefore, the vector position of the center of mass at t = 2.80 is -0.70 cm.