## College Physics (7th Edition)

a). $513N$ b). $0.151m/s$
a). Change in velocity $=47+122=169mi/hr=75.5m/s$ Therefore, $F=ma=m\frac{\Delta v}{\Delta t}=0.17\frac{75.5}{0.025}=513N$ b). $v=a_{g}t=\frac{F}{m_{g}}t=\frac{513}{85}0.025=0.151m/s$