# Vectors and Kinematics

1. Feb 14, 2010

1. The problem statement, all variables and given/known data
Jordyn throws an eraser at her teacher, it moves horizontally at 8m/s when the teacher smacks it with a metrestick. The eraser leaves the metrestick horizontally at 15 m/s [N30W]. It is in contact with the stick for 0.03s. Determine the average acceleration of the eraser while being hit by the metrestick. Draw a vector diagram showing the motion.

2. Relevant equations
We were not given specific equations for this question, but since we are given two velocities, time, and are asked for average acceleration, i figured that maybe a=(V1-V2)/t might work.

3. The attempt at a solution
So, first i thought that i could use the equation above, and just plug in the given information, but then i figured that the eraser stopped when it hit the stick. So, it would go from 8m/s to 0m/s and then to 15m/s. I then used two equations (both being a=(V1-V2)/t). For the first, i used V1=8m/s and V2=0m/s; for the second, i used V1=0m/s and V2=15m/s. Since we have to find the average acceleration, i thought maybe i could average out both answers from the two equations. I doubt this works though, maybe it does, i'm not sure. It seems too simple. Especially because i dont know if it's right to use the same time for both equations. Also, i have absolutely no idea how this relates to vectors. If i had to add/subtract both vectors then i could have just drawn that, but they're two seperate things. Please help and thanks so much!

2. Feb 15, 2010

### rallizes

i didn't read your attempt cause im really sleepy and the paragraph is really big but

you want to say the eraser went from 8m/s to 15m/s in the other direction in .3 seconds.

so use kinematics equation vfinal=vinitial+at, so vfinal=-vinitial+a(t), a=(vfinal+vinitial)/(time). vinitial is the opposite sign of vfinal as it is in the opposite direction. you might have that written down i dunno. don't worry about the instantaneous velocity of the eraser cause it asks for the average acceleration, i confused myself the other day cause i thought another problem implied infinite acceleration by saying something stopped right away

3. Feb 15, 2010