Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 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? Hint distance A to B = B to A Answer (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