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Larry leaves home at 2:08 and runs at a constant speed to the lamppost. He reaches the lamppost at 2:15, immediately turns, and...Asked by Vishnu
Larry leaves home at 2:08 and runs at a constant speed to the lamppost. He reaches the lamppost at 2:15, immediately turns, and runs to the tree. Larry arrives at the tree at 2:29. What is Larry's average velocity during his trip from home to the lamppost, if the lamppost is 308.0 m west of home, and the tree is 688.0 m east of home?
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velocity is change of position (a vector) per time.
change of position: final position-initial
= 688E+308W=380W
avgvelocity=380W/total time.
I got
avgvelocity = 380W/ total time
= 380 / 21 seconds
= 18.1 m/s
Why is my answer incorrect?
------------------
velocity is change of position (a vector) per time.
change of position: final position-initial
= 688E+308W=380W
avgvelocity=380W/total time.
I got
avgvelocity = 380W/ total time
= 380 / 21 seconds
= 18.1 m/s
Why is my answer incorrect?
Answers
Answered by
bobpursley
Rereading. The lamppost is 308m west of home.
velocity=308/time
velocity=308/time
Answered by
Vishnu
308/21 = 14.66 m/s
I still get a wrong answer
I still get a wrong answer
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