At one instant a bicyclist is 47.0 m due east of a park\'s flagpole, going due s

Question

At one instant a bicyclist is 47.0 m due east of a park's flagpole, going due south with a speed of 14.0 m/s. Then 22.0 s ater, the cyclist is 47.0 m du north of the flagole, goin due east with a speed of 14.0 m/s. For the cyclist in this 22.0 s interval, what are the a) magnitude and b) direction of the displacement, the c) magnitude and d) direction of the average velocity, and the e) magnitude and f) direction of the average acceleration? (give all directions as positive angles realative to due east, where positive is measured going counterclockwise).

Explanation / Answer

taking the flagpole to be the origin

x= 47i
y=47j
displacement= -47i + 47 j
magnitude= 472= 66.47

direction north west

time= 22 sec

hence average velocity= displacement/time= (47/22) (-i+j)

direction is along the direction of displacement, NW

accelaration= final vel-initial vel/time= 14i - (-14j) / 22 = 0.636 (i+j)

mag= 0.9

direction= north east

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