Answer
a) $-1.25\times10^5N$
b) $8.5 g's$
Work Step by Step
a) The average force is the momentum change divided by the elapsed time.
$$F_{avg}=\frac{\Delta p}{\Delta t}=\frac{m\Delta v}{\Delta t}=\frac{(1500kg)(0-45km/h)(\frac{1m/s}{3.6km/h})}{0.15s}=-1.25\times10^5N$$
b) Use the $F=ma$
$$F_{avg}=ma_{avg}$$
$$a_{avg}=\frac{F_{avg}}{m}=\frac{1.25\times10^5N}{1500kg}=83.33m/s^2(\frac{1g}{9.8m/s^2})\approx8.5g's$$
