Question

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

Answer 1

`\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)((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}`