Post a New Question


posted by on .

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?


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?


physics - bobpursley, Wednesday, October 8, 2008 at 7:09am
Rereading. The lamppost is 308m west of home.


physics - Vishnu, Wednesday, October 8, 2008 at 9:28am
308/21 = 14.66 m/s

I still get a wrong answer

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question