Physics I help urgently needed please! Velocity

A dog running in an open field has components of velocity Vx= 1.8m/s and Vy=-2.5m/s at time t1= 11.9s. For the time interval from t1= 11.9s to t2= 24.5s, the average acceleration of the dog has magnitude 0.53m/s^2 and direction 31.5 degrees measured from the x-axis toward the y-axis.

A) At time t2 = 24.5 s, what are the x- and y-components of the dog's velocity? (m/s)
B)What is the magnitude of the dog's velocity? (m/s)
C)What is the direction of the dog's velocity (measured from the x-axis toward the y-axis)? (in degrees)
Please explain any help is greatly appreciated!

You will have to break up the given acceleration into components and then write the components in terms of the velocity change;
ax = aCos(31.5) = (Vxf - Vxi)/(tf - ti)

solve for Vxf
Vxf = a(tf-ti)Cos(31.5) + Vxi
Vxf = (.53)(12.6)Cos(31.5) + 1.8

Similarly for the y-component;
ay = aSin(31.5) = (Vyf - Vyi)/(tf - ti)
Vyf = a(tf-ti)Sin(31.5) + Vyi
Vyf = (.53)(12.6)Sin(31.5) - 2.5

These are the velocity components at the end of the time interval (at time tf=24.5s)

Once you find the numbers for Vxf & Vyf , the magnitude of the velocity Vf is;
Vf = SqRt[Vxf^2 + Vyf^2]

The direction measured from the x-axis is found from;
Tan() = Vyf/Vxf