At one instant a bicyclist is 50 m due east of a park\'s flagpole, going duesout

Question

At one instant a bicyclist is 50 m due east of a park's flagpole, going duesouth with a speed of 20 m/s.Then, 8 s later, the cyclistis 30 m due north of theflagpole, going due east with a speed of 11 m/s. For the cyclist in this 8 s interval, find each of the following. (a) displacement magnitude m direction °north of west
(b) average velocity magnitude m/s direction °north of west
(c) average acceleration magnitude m/s2 direction °north of east (a) displacement magnitude m direction °north of west
(b) average velocity magnitude m/s direction °north of west
(c) average acceleration magnitude m/s2 direction °north of east magnitude m direction °north of west

Explanation / Answer

For starters, we know that the average displacement is just thedistance that he is from the starting point divided by thetime. So, the average displacement D= (50^2 +30^2)/8s The direction is going to be = 180 - arctan(y/x)   y = 30, x = -50 The 180 is in there because it's "north of west" which is thesecond quadrant. Essentially , you do the same exact thing for velocity as you didfor displacement. To find the magnitude of the average acceleration, I would splitthe velocity vector into x and y coordinates. Subtract the finalvelocities from the initial velocities. and then divide thatdifference by time. after you've done that, reassemble the twovelocity vectors into one magnitude using the pythagoreanthm.

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