Answer
The average force that acts on the car during the turn is directed at an angle of $~~45^{\circ}~~$ below the positive x axis.
Work Step by Step
In part (a), we found that the impulse on the car due to the turn is $~~J=(7420~\hat{i}-7420~\hat{j})~kg~m/s$
We can find the average force that acts on the car during the turn:
$F~t = J$
$F = \frac{J}{t}$
$F = \frac{(7420~\hat{i}-7420~\hat{j})~kg~m/s}{4.6~s}$
$F = (1613~\hat{i}-1613~\hat{j})~N$
We can find the angle (below the positive x axis) of the average force that acts on the car during the turn:
$tan~\theta = \frac{1613~N}{1613~N}$
$\theta = tan^{-1}~(\frac{1613~N}{1613~N})$
$\theta = tan^{-1}~(1)$
$\theta = 45^{\circ}$
The average force that acts on the car during the turn is directed at an angle of $~~45^{\circ}~~$ below the positive x axis.
