Larry leaves home at 2:08 and runs at a constant speed to the lamppost. He reaches the lamppost at 2:12, immediately turns, and runs to the tree. Larry arrives at the tree at 2:28. What is Larry's average velocity during his trip from home to the lamppost, if the lamppost is 321.0m west of home, and the tree is 698.0m east of home?
b). What is Larry's average velocity during his trip from the lamppost to the tree?
c). What is the average velocity for Larry's entire run?
Physics - Henry, Sunday, September 30, 2012 at 7:03pm
a. V=d/t = 321/(2:12-2:08)=321/4min =
80.25 m/min = 80.25m/60s = 1.34 m/s.
b. V = d/t = (698+321)/(2:28-2:12) =
1019/16 = 63.7m/min = 63.7/60s=1.06 m/s
c. V = (698+321)/(2:28-2:08) = 1019/20 =
50.95m/min = 50.95m/60s = 0.849 m/s.
Physics - Anonymous, Monday, February 1, 2016 at 3:50pm