## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $v_{Tg}$ be Trent's velocity with respect to the ground. Let $v_{Sg}$ be Susan's velocity with respect to the ground. Let $v_{TS}$ be Trent's velocity with respect to Susan. $v_{Tg} = v_{TS}+v_{Sg}$ $v_{TS} = v_{Tg} - v_{Sg}$ We can find the east component of $v_{TS}$: $v_{TS,east} = 45~mph - 0 = 45~mph$ (east) We can find the south component of $v_{TS}$: $v_{TS,south} = 0 - (-60~mph) = 60~mph$ (south) We can find Trent's speed relative to Susan's reference frame: $v_{TS} = \sqrt{(45~mph)^2+(60~mph)^2}$ $v_{TS} = 75~mph$ Trent's speed is 75 mph relative to Susan's reference frame.