(i) Sketch:

car racetrack

Note the axes are x- and y-based, and do not involve t.

(ii) Acceleration:

Horizontal acceleration:

x = 20 + 0.2t3

`v_x=(dx)/(dt)=0.6t^2`

`a_x=(d^2x)/(dt^2)=1.2t`

At t = 3.0, ax = 3.6

Vertical acceleration:

y = 20t − 2t2

`v_y=(dy)/(dt)=20-4t`

`a_y=(d^2y)/(dt^2)=-4`

At t = 3.0, ay = -4

Now

`a=sqrt((a_x)^2+(a_y)^2)=sqrt(3.6^2+(-4)^2)=5.38`

and

`theta_a=arctan((a_y)/(a_x))=arctan\ ((-4)/3.6)=312^"o"` [4th quadrant]

So the car's acceleration has magnitude 5.38 ms-2, and direction `312^@` from the positive x-axis.