Initial velocity is v0=5.0km/h=1.389m/sv_0=5.0 km/h= 1.389 m/sv0=5.0km/h=1.389m/s
Final velocity is v=0v=0v=0
Distance traveled is d=2.0md=2.0 md=2.0m
The acceleration is
a=v2−v022d=(0)2−(1.389)22(2.0)=−0.482m/s2a=\frac{v^2-v_0^2}{2d}=\frac{(0)^2-(1.389)^2}{2(2.0)}=-0.482 m/s^2a=2dv2−v02=2(2.0)(0)2−(1.389)2=−0.482m/s2
the magnitude of the net force on the driver is
F=ma=(80kg)(0.482m/s2)=38.58N=38.6NF=ma=(80 kg)(0.482 m/s^2)=38.58 N=38.6 NF=ma=(80kg)(0.482m/s2)=38.58N=38.6N
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