At one instant a bicyclist is 75 m dueeast of a park\'s flagpole, going due sout

Question

At one instant a bicyclist is 75 m dueeast of a park's flagpole, going due south with a speed of11 m/s. Then, 22 s later, the cyclist is 25 m due north of the flagpole, going due eastwith a speed of 14 m/s. For thecyclist in this 22 s interval, findeach of the following. (a) displacement magnitude 1 m direction 2° north of west
(b) average velocity magnitude 3 m/s direction 4° north of west
(c) average acceleration magnitude 5 m/s2 direction 6° north of east (a) displacement magnitude 1 m direction 2° north of west
(b) average velocity magnitude 3 m/s direction 4° north of west
(c) average acceleration magnitude 5 m/s2 direction 6° north of east magnitude 1 m direction 2° north of west

Explanation / Answer

(a)   Initial position is 75m due east.          thereforex1 = (75m)i       Final position is 25m duenorth          thereforex2 = (25m)j       then displacement is x =x2 - x1                                        = [(25m)j] - [(75m)i]                                        = (-75m)i +(25m)j       then the magnitude ofdisplacement is [(-75m)2 + (25m)2] =79m       direction istan-1(25/-75)toward north of east                      = tan-1(25/75)toward north of west                       =18.43o toward north of west (b)   Initial velocity is 11m/stoward south       thereforev1 = -(11m/s)j       Final velocity is 14m/stoward east.       thereforev2 = (14m/s)i       then average velocity isvav = [v1 +v2]/2                                                 =[-(11m/s)j + (14m/s)i]/2                                                 =(7m/s)i +(-5.5m/s)j       then theaverage velocity is vav =[(7m/s)2 + (-5.5m/s)2]                                                     = 8.9m/s       Direction of averagevelocity is tan-1[(-5.5m/s)/(7m/s)] toward north ofeast                                                     =  tan-1[(5.5m/s)/(7m/s)]toward north of west                                                    = 38otoward north of west (c)   Average acceleration = (change invelocity) / (time)                                        = (v2 - v1)/t                                        = [(14m/s)i -(-11m/s)j]/(22s)                                        = [((14/22)m/s2)i+((11/22)m/s)j ]       then the magnitude ofacceleration [(14/22)2 +(11/22)2]m/s2 = 17.8m/s2       Direction of accelerationis tan-1(11/14) toward north of east = 38.15otoward north of east             (b)   Initial velocity is 11m/stoward south       thereforev1 = -(11m/s)j       Final velocity is 14m/stoward east.       thereforev2 = (14m/s)i       then average velocity isvav = [v1 +v2]/2                                                 =[-(11m/s)j + (14m/s)i]/2                                                 =(7m/s)i +(-5.5m/s)j       then theaverage velocity is vav =[(7m/s)2 + (-5.5m/s)2]                                                     = 8.9m/s       Direction of averagevelocity is tan-1[(-5.5m/s)/(7m/s)] toward north ofeast                                                     =  tan-1[(5.5m/s)/(7m/s)]toward north of west                                                    = 38otoward north of west (c)   Average acceleration = (change invelocity) / (time)                                        = (v2 - v1)/t                                        = [(14m/s)i -(-11m/s)j]/(22s)                                        = [((14/22)m/s2)i+((11/22)m/s)j ]       then the magnitude ofacceleration [(14/22)2 +(11/22)2]m/s2 = 17.8m/s2       Direction of accelerationis tan-1(11/14) toward north of east = 38.15otoward north of east                   then theaverage velocity is vav =[(7m/s)2 + (-5.5m/s)2]                                                     = 8.9m/s       Direction of averagevelocity is tan-1[(-5.5m/s)/(7m/s)] toward north ofeast                                                     =  tan-1[(5.5m/s)/(7m/s)]toward north of west                                                    = 38otoward north of west (c)   Average acceleration = (change invelocity) / (time)                                        = (v2 - v1)/t                                        = [(14m/s)i -(-11m/s)j]/(22s)                                        = [((14/22)m/s2)i+((11/22)m/s)j ]       then the magnitude ofacceleration [(14/22)2 +(11/22)2]m/s2 = 17.8m/s2       Direction of accelerationis tan-1(11/14) toward north of east = 38.15otoward north of east       Direction of accelerationis tan-1(11/14) toward north of east = 38.15otoward north of east            

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