Connect with knowledgeable individuals and find the best answers at IDNLearn.com. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.

I really need help with this question-only reply if you know how to solve it please

I Really Need Help With This Questiononly Reply If You Know How To Solve It Please class=

Sagot :

[tex]\begin{gathered} a)\text{ For a double}.\text{ The end of this }run\text{ is point A} \\ \text{Then, } \\ x-\text{displa}cement=27.4\text{ m} \\ y-\text{displa}cement=27.4\text{ m} \\ |A|=\sqrt{(27.4\text{ m})^2+(27.4\text{ m})^2} \\ |A|=38.75m \\ \theta=\tan ^{-1}(\frac{27.4m}{27.4m}) \\ \theta=45\text{ \degree} \\ \text{The magnitude is }38.75m\text{ and the angle is 45\degree} \\ \\ b)\text{For a triple}.\text{ The end of this }run\text{ is point }B \\ \text{The runner has to reach first base, then second base and finally thrid base} \\ x-displacement=\text{first base+thrid base} \\ x-\text{displacement =}27.4m-27.4\text{ m} \\ x-\text{displacement =0 m} \\ y-\text{displacement =second base} \\ y-\text{displacement =27.4m} \\ |B|=\sqrt{(\text{0 m})^2+(\text{27.4m})^2} \\ |B|=27.4m \\ There\text{ is only y- displacement, positive }y\text{ axi}s \\ \text{Then} \\ \theta=90\text{ \degree} \\ \text{The magnitud is 27.4 m and the angle is 90\degree} \end{gathered}[/tex]

We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.