avgˆˆ(483 km)i (966 km)j(215 km/h)i (429 km/h)j.2.25 hv−==−Gwith a magnitude 22||(215 km/h)( 429 km/h)480 km/h.v=+−=G(d) The direction of vGis 26.6° east of south, same as in part (b). In magnitude-angle notation, we would have (480 km/h 63.4 ).v=∠−°G(e) Assuming the ABtrip was a straight one, and similarly for the BCtrip, then ||GrABis the distance traveled during the ABtrip, and GrBCis the distance traveled during the BCtrip. Since the average speed is the total distance divided by the total time, it equals 483 km 966 km644 km/h.2.25 h+=8. Our coordinate system has #ipointed east and
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