91Ó°ÊÓ

Susan is driving north at $60\text{mph}$and Trent is driving east at $45\text{mph}$, and they are approaching an intersection.

The following figure shows the vector representation of the speed of the Susan with respect to Trent's reference.

The speed of the Trent with respect to the speed of the Susan's reference frame, from the shown figure $\left(I\right)$is,

${\left({\stackrel{→}{V}}_{TS}\right)}^2={\left({\stackrel{→}{V}}_{SG}\right)}^2+{\left({\stackrel{→}{V}}_{TG}\right)}^2$

Here, ${\stackrel{→}{V}}_{TS}$is the speed of the Trent with respect to the speed of the Susan's reference frame, ${\stackrel{→}{V}}_{SG}$is the speed of the Susan with respect to the ground, and ${\stackrel{→}{V}}_{TG}$is the speed of the Trent with respect to ground.

role="math" localid="1652321923075" $\begin{array}{l}{\left({\stackrel{→}{V}}_{TS}\right)}^2={\left(60\text{″¾±è³ó}\right)}^2+{\left(45\text{″¾±è³ó}\right)}^2\\ {\stackrel{→}{V}}_{TS}=\sqrt{5625{\text{mph}}^2}=75\text{mph}\end{array}$

Thus, the Trent's speed relative to Susan's reference frame is$75\text{″¾±è³ó}$ .