I'm stuck at part b on this problem. ive been working at it for an hour at least! lol(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

At one instant a bicyclist is 50 m due east of a park's flagpole, going due south with a speed of 7 m/s. Then, 24 s later, the cyclist is 70 m due north of the flagpole, going due east with a speed of 20 m/s. For the cyclist in this 24 s interval, find each of the following.

(a) displacement

magnitude: ? m

direction: ? mark° north of due west

(b) average velocity

magnitude: ? m/s

direction: ? ° north of due west

(c) average acceleration

magnitude: ? m/s2

direction: ? ° north of due east

2. Relevant equations

[tex]\vec{}r[/tex] = [tex]\sqrt{}((r\hat{i})^{2} + (r\hat{j})^{2})[/tex]

3. The attempt at a solution

alright so for part (a) i used 50\hat{i} and 70\hat{j} in the equation above to get 86.023m

then inversetan (70/50) to get 54.462 degrees. i thought that 54.462 was the angle for norht of due east, but its the correct answer. It would be nice if someone could explain that to me.

then with part (b) im completely lost, i tried plugging in the 7m/s and the 20m/s the same way as i did for the displacement but got 21.190 degrees. thats wrong and now iv been stuck forever it seems. any help would be much appreciated. thanks!

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# Homework Help: Really fast bicyclist in the park

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