A car makes a trip due west for three-fourths of the time anddue east one-fourth

Question

A car makes a trip due west for three-fourths of the time anddue east one-fourth of the time. The average westward velocity hasa magnitude of 25 m/s, and the average eastward velocity has amagnitude of 17 m/s. What is the magnitude and direction, for theentire trip?
This is how I've set it up: Average velocity west: = -25 m/s = (-18.75 m) / (3/4 s) Average velocity east: = 17 m/s = (4.25 m) / (1/4 s) Someone showed me their calculations and thought it was -14.5m/s but that is not correct. A car makes a trip due west for three-fourths of the time anddue east one-fourth of the time. The average westward velocity hasa magnitude of 25 m/s, and the average eastward velocity has amagnitude of 17 m/s. What is the magnitude and direction, for theentire trip?
This is how I've set it up: Average velocity west: = -25 m/s = (-18.75 m) / (3/4 s) Average velocity east: = 17 m/s = (4.25 m) / (1/4 s) Someone showed me their calculations and thought it was -14.5m/s but that is not correct.

Explanation / Answer

      Let total time = tseconds      average eastward velocity = 17m/s      average westward velocity = -25m/s (here westward velocity is taken as negative because itis opposite to east)      time of travel during east = t /4      time of travel during east = 3 t /4      displacement along east = velocity *time = ( 17 m/s)(t / 4)      displacement along west =velocity * time = (-25 m/s)(3 t / 4)      Average velocity of the entiretrip = total displacement / total time taken                                                         = ( 17 m/s)(t / 4) + (-25 m/s)(3 t / 4) / t                                                         = (17 - 75) / 4                                                          = -14.5 m/s      direction of average velocity =west

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