Physics Help Forum [SOLVED] Finding position, average velocity...

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Sep 13th 2010, 06:47 PM #1 Member   Join Date: Sep 2010 Posts: 30 [SOLVED] Finding position, average velocity... So i have found velocity at 4 seconds, which gave me 8 m/s. How could i find position after the first 7 seconds? I got 2m/s but it was wrong =( Also on part 3 and 4 i would like help also. Thank you for taking your time in reading and helping me, this is giving me problems.
 Sep 14th 2010, 12:11 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,322 Part 1 is correct. The area under the acceleratoin curve gives you the velocity, and that is 2s x 4m/s^2 = 8m/s For part 2, one way to do this is to plot the velocity graph versus time, and then calculate the area under it. The velocity graph is a series of straight line sehgments: a) From 0 second to 1 second v = 0. The area under that line is obviously 0, meaning the particle didn't move at all for the first 1 second. b) From 1 second to 3 seconds the velocity increases in a straight line from 0 to 8 m/s. The area under the velocity curve is a triangle of area 1/2 x 2s x 8 m/s = 8 m. c) From 3 seconds to 4 seconds the velocity is constant 8 m/s. So the area under that portion of the velocity graph is 1s x 8m/s = 8 m. d) From 4 seconds to 7 seconds the velocity decreases in a straight line from 8 m/s to 0m/s. The area under the curve is anothetr triangle, this time of area 1/2 x 8 m/s x 3 s = 12 m. If you add these segments up, what do you get for the total displacement? 3. The velocity from t=7 seconds to 9 seconds is 0, so how far does the particle travel during that time? 4. The average veociity is the total distance traveled between t= 1 and t=9 (from part 2 + part 3) divided by the total time, which is 8 seconds.
 Sep 14th 2010, 12:50 PM #3 Member   Join Date: Sep 2010 Posts: 30 Ok so my position would be be 35 meters at 7 seconds? Since i added 8+8+15+ = 31 meters? For number 3, from 7 to 9 seconds the velocity is 0, so that mean it travels 4 meters? And for number 4, the formula for average velocity is (X final - X initial)/ (t final - t initial) = (35/8) = 4.375? It's displacement divided by time Is that correct? Last edited by rodjav305; Sep 14th 2010 at 01:01 PM. Reason: mistake on calculation
 Sep 14th 2010, 01:02 PM #4 Member   Join Date: Sep 2010 Posts: 30 Thank you so much for your help!!! I can't believe i couldn't do this simple problem but i have no trouble on the harder ones =/ Thank you again!!
 Sep 14th 2010, 07:15 PM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,322 rodjav: I apologize, but I realize I made an error in my earlier post. The velocity at t = 7s = 2 m/s, not 0 as I had before. This is because the change in velocity from t=4 to t = 7 is -6m/s. Hence the distance traveled in that segment is 1/2(8 + 2)3 = 15 m, and the position at t=7 is 8m + 8m + 15m = 31m. From t=7 to t=9 the velocity is 2 m/s, so it travels another 4 m in those 2 seconds. Hence at t=9 the particle is at x=35m, and its average velocity during the 8 seconds form t=1 to t = 9 is 35m/8s = 4.375m/s. Hope this clears it up for you!

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