Unformatted text preview: 17. We find the time for the outgoing 200 km from t 1 = d 1 / v 1 = (200 km)/(90 km/h) = 2.22 h. We find the time for the return 200 km from t 2 = d 2 / v 2 = (200 km)/(50 km/h) = 4.00 h. We find the average speed from average speed = ( d 1 + d 2 )/( t 1 + t lunch + t 2 ) = (200 km + 200 km)/(2.22 h + 1.00 h + 4.00 h) = 55 km/h . Because the trip finishes at the starting point, there is no displacement; thus the average velocity is æ = ∆ x / ∆ t = . 18. If v AG is the velocity of the automobile with respect to the ground, v TG the velocity of the train with respect to the ground, and v AT the velocity of the automobile with respect to the train, then v AT = v AG – v TG . If we use a coordinate system in the reference frame of the train with the origin at the back of the train, we find the...
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- Fall '08
- 1g, 4.3 m/s, 2 1G, 6.6 min, 9.9 km