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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`
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`
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