31. We choose the positive direction to be that of the initial velocity of the car (implying thata<0 sinceit is slowing down). We assume the acceleration is constant and use Table 2-1.(a) Substitutingv0= 137 km/h=38.1m/s,v=90km/h=25m/s, anda=−5.2m/s2intov=v0+at,we obtaint=25 m/s−38 m/s−5./s2=2.5s.(b) Wetakethecartobeatx= 0 when the brakes are applied(at timet= 0). Thus, thecoordinate of the car as afunction of time is given byx= (38)t+12(−5.2)t2in SI units. This function
This is the end of the preview.
access the rest of the document.