SOLUTION: A car travels along a straight line at a constant speed of 55.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average veloc
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Question 494438: A car travels along a straight line at a constant speed of 55.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 26.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
(c) What is the average speed for this new trip? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A car travels along a straight line at a constant speed of 55.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 26.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
Avg speed = 2*r1*r2/(r1 + r2) (I figured that out some time ago0
26 = 2*55*r2/(55+r2)
715 + 13*r2 = 55*r2
715 = 42*r2
r2 = 715/42 mi/hr =~ 17 mi/hr
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(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
If the distance is the same, the direction is irrelevant.
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(c) What is the average speed for this new trip?
What new trip? That's already answered ??
