1. A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. (a) What is her average speed over the entire trip? (b) What is her average velocity over the entire trip?

distance `A to B = B to A`
(a) speed: `3.75 m//s` (b) velocity: `0 m//s`
Show Steps
Given: `v_(AB) = 5.00 m//s` `v_(BA) = 3.00 m//s` distance: `d` Calculations: (a) `v_a = (2d)/(t_(AB)+t(BA))` `t_(AB) = d/v_(AB)` and `t_(BA) = d/v_(BA)` `=> v_a =(2d)/(d/v_(AB) + d/v_(BA))` `=> v_a =(2d)/((1/v_(AB) + 1/v_(BA))d)` `=> v_a =(2)/((1/v_(AB) + 1/v_(BA)))` `=> v_a =(2)(v_(AB)v_(BA))/(v_(AB) + v_(BA))` `v_a =(2)((5.00 m//s)(3.00m//s))/((5.00 m//s)+ (3.00 m//s))` `v_a = 3.75 m//s` (b) `v_a = (∆x)/(∆t)` Since the person returns to the starting point: `∆x = 0` and `v_a = 0`