Constant acceleration problem (r = (3.2t + 1.8t^2) i + (1.7t - 2.4t^2) j m)

my problem is:
The position of an object as a function of time is given by
r = (3.2t + 1.8t^2) i + (1.7t - 2.4t^2) j
(the "r" has a vector arrow above it and the "i" and "j" have "hats" above them)
where "t" is the time in seconds. What are the magnitude and direction of the acceleration?

i'm just so confused as to what i'm supposed to do. the answer is 6.0 m/s^2 and 53 degrees below the x-axis. i've tried taking the second derivatives and using the formulas i have but since i'm not given time and i need to find acceleration, i'm not sure how to approach the problem. a push in the right direction would be much appreciated. thank you

you can do it by second derivatives..

first derivative= (3.6t + 3.2) i + (-4.8t+1.7) j

second derivative = (3.6)i + (-4.8)j

hence, acceleration = under root[3.6^2 + (-4.8)^2] = 6

for angle, tan(angle) = -4.8/3.6

hence, angle = tan inverse(-4.8/3.6) = -53.13

The second derivative of this position vector does not depend on time, that's why it is called a constant acceleration problem.....

a = d² r / dt² = 3.6 î - 4.8 ĵ

so the magnitude of a is

√( (3.6)^2 + (-4.8)^2 ) = 6.0

and for the angle with the î direction:

tan(α) = -4.8/3.6 => α = -53 degrees