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)The displacement of the bicyclist during the 22 s intervalis S = 75i + 25j The magnitude of the displacement is |S| = [(75)2 +(25)2]1/2 or |S| = [5625 + 625]1/2 = 79 m The direction of the displacement is tan = (Sy/Sx) = (25/75) =(1/3) or = tan-1(0.3333) = 18.4o northof west (b)The average velocity is v = -11i + 14j The magnitude of the average velocity is |v| = [(-11)2 + (14)2]1/2 =17.8 m/s The direction of the average velocity is tan1 = (vy/vx) =(14/-11) or 1 = tan-1(-14/11) =-51.8o north of west (c)The acceleration of the bicyclist in the first caseis vy2 - uy2 =2axSx or ax = (vy2 -uy2/2Sx) vy = -11 m/s,uy = 0 and Sx =75 m The acceleration of the bicyclist in the second caseis vx2 - ux2 =2aySy or ay = (vx2 -ux2/Sy) vx = 14 m/s,ux = 0 and Sy =25 m The acceleration is a = iax + jay The magnitude of the average acceleration is |a| = [ax2 +ay2]1/2 The direction of the average acceleration is tan2 =(ay/ax) north of east or ax = (vy2 -uy2/2Sx) vy = -11 m/s,uy = 0 and Sx =75 m The acceleration of the bicyclist in the second caseis vx2 - ux2 =2aySy or ay = (vx2 -ux2/Sy) vx = 14 m/s,ux = 0 and Sy =25 m The acceleration is a = iax + jay The magnitude of the average acceleration is |a| = [ax2 +ay2]1/2 The direction of the average acceleration is tan2 =(ay/ax) north of east The direction of the average acceleration is tan2 =(ay/ax) north of east

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