SOLUTION: A locomotive travels on a straight track at a constant speed of 40 mi/h, then reverses direction and returns to its starting point, traveling at a constant speed of 60 mi/h.
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Question 653001: A locomotive travels on a straight track at a constant speed of 40 mi/h, then reverses direction and returns to its starting point, traveling at a constant speed of 60 mi/h.
What is the average velocity for the round-trip?
What is the average speed for the round-trip?
What is the total distance traveled by the train if the total trip took 5 h?
I have no idea where to even begin...
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let the direction of travel for the initial leg of the trip be 
. Then the direction of travel for the return leg is \ \text{mod}\ 360^\circ)
. The vector sum of the two opposing vectors is the difference of the magnitudes in the direction of the greater magnitude, i.e. 20 mph @ .
The average speed is the total distance traveled divided by the total travel time. The time for the outward leg is:
hours.
Where the one-way distance is 
The time for the return leg is:
hours.
The sum of these times is the total travel time:
hours.
The total distance travelled is 
, so the average speed is:
mph.
If the trip took 5 hours...
If you go 48 mph for 5 hours, how far did you go?
John
My calculator said it, I believe it, that settles it
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