Answer
The magnitude of the average force that acts on the car during the turn is $~~2280~N$
Work Step by Step
In part (a), we found that the impulse on the car due to the turn is $~~(7420~\hat{i}-7420~\hat{j})~kg~m/s$
We can find the magnitude of the impulse:
$J = \sqrt{(7420~kg~m/s)^2+(-7420~kg~m/s)^2}$
$J = 10,493~kg~m/s$
We can find the magnitude of the average force that acts on the car during the turn:
$F~t = J$
$F = \frac{J}{t}$
$F = \frac{10,493~kg~m/s}{4.6~s}$
$F = 2280~N$
The magnitude of the average force that acts on the car during the turn is $~~2280~N$
