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?

Rereading. The lamppost is 308m west of home.

velocity=308/time

308/21 = 14.66 m/s

I still get a wrong answer

Your approach and calculation for finding the average velocity is correct. However, there seems to be an error in the calculation of the total time. Let's correct it:

To find the total time of Larry's trip, we need to consider the time it took for him to reach the lamppost and the time it took for him to reach the tree.

Time from home to lamppost: 2:15 - 2:08 = 7 minutes = 7 * 60 seconds = 420 seconds
Time from lamppost to tree: 2:29 - 2:15 = 14 minutes = 14 * 60 seconds = 840 seconds

Total time = time home to lamppost + time lamppost to tree = 420 seconds + 840 seconds = 1260 seconds

Now let's calculate the average velocity:

average velocity = change in position / total time
= 380W / 1260 seconds
= 0.301 W m/s (rounded to 3 decimal places)

So, Larry's average velocity during his trip from home to the lamppost is approximately 0.301 m/s west.