26. The stopping force GFand the path of the passenger are horizontal. Our +xaxis is in the direction of the passenger’s motion, so that the passenger’s acceleration (‘‘deceleration” ) is negative-valued and the stopping force is in the –xdirection: ˆiFF=−G. Using Eq. 2-16 withv0= (53 km/h)(1000 m/km)/(3600 s/h) = 14.7 m/s
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