A baseball “diamond” (Figure) is a square with sides 90 ft

Step 1 of 3

a.) We have to find the displacement vector of a base runner who has just hit a double.

The position of the base runner who has just hit a double on the baseball diamond (square) is at A(90,90) as shown in the figure below. 

 Thus, the displacement vector of the base runner who is at A is given by

\(\begin{aligned} \vec{A} & =\left(\mathrm{A}_{\mathrm{x}}\right) \hat{\boldsymbol{x}}+\left(\mathrm{A}_{\mathrm{y}}\right) \hat{\boldsymbol{y}} \\ & =(90 \mathrm{ft}) \hat{\boldsymbol{x}}+(90 \mathrm{ft}) \hat{\boldsymbol{y}} \end{aligned}\)

Therefore, the displacement vector of a base runner who has just hit a double is \(\vec{A}=(90 \mathrm{ft}) \hat{x}+(90 \mathrm{ft}) \hat{y}\).