#$&* course phy201 10-30-10 006. `query 6*********************************************
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Given Solution: The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions: • the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities justification in terms of definitions: if acceleration is uniform, then the v vs t graph is linear so that the average velocity is equal to the average of initial and final velocities • multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m justification in terms of definitions: ave velocity is ave rate of change of position with respect to clock time, so vAve = `ds / `dt, from which it follows that `ds = vAve * `dt • the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2 justification in terms of definitions: ave acceleration is ave rate of change of velocity with respect to clock time, so aAve = `dv / `dt. in this situation acceleration is uniform, so we can if we wish use just plain a instead of aAve The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters. A solution to the problem: Using the fourth equation of motion with the given information (`ds, a and vf) we have vf^2 = v0^2 + 2 a `ds , which we solve for v0 to get v0 = +- sqrt( vf^2 - 2 a `ds) = +- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) = +- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) = +- sqrt( 400 m^2 / s^2) = +- 20 m/s. If vf = 20 m/s then we could directly reason out the rest (vAve would be 25 m/s, so it would take 5 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - 20 m/s) / (2 m/s^2) = 10 m/s / (2 m/s^2) = 5 s. If vf = -20 m/s then we could directly reason out the rest (vAve would be (-20 m/s + 30 m/s) / 2 = 5 m/s, so it would take 25 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - (-20 m/s) ) / (2 m/s^2) = 50 m/s / (2 m/s^2) = 25 s. STUDENT COMMENT Umm, evidently I did NOT understand the problem.. even looking back, I’m still not sure how everything in the question exactly relates to the answer. I understand the given answer and what it means, but the original questions are very confusing to me! INSTRUCTOR RESPONSE The point is that the student's solutions are inconsistent. Using the initial velocity obtained by the student, the change in velocity would be 30 m/s and the acceleration would be 3 m/s^2. This of course contradicts the given acceleration, which was 2 m/s^2. So the student's solution contradicts the given information. STUDENT COMMENT I do not understand this problem at all. The information given all makes sense and I know that I am looking for some way to make it all fit together to verify the students results. I understand it a little better and understand the path I was supposed to take after reading through the explanation, but am going to have to work on problems like these. INSTRUCTOR RESPONSE It would be really beneficial for you to go through the given solution one phrase at a time, and tell me exactly what you do and do not understand. For example in the first few lines: The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions: • the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities **** do you understand how the given information leads to the conclusion that the average velocity is 15 m/s? **** **** what do you and do you not understand about the meaning of the statement 'since acceleration is uniform is simply the average of the initial and final velocities'? **** the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s **** what do you and do you not understand about the details of this statement and its overall meaning? **** so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2 **** what do you and do you not understand about the meaning of the statement acceleration is `dv / `dt? **** **** what do you and do you not understand about why in this case `dv is 30 m/s and `dt is 10 s? **** **** what do you and do you not understand about the calculation 30 m/s / (10 s) = 3 m/s^2? **** You should deconstruct the entire solution, one phrase at a time, and tell me what you do and do not understand about each. With that information I can help you address the things you don't understand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I have a better understand of what exactly you were looking for now, ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance. If the positive direction is down the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the air resistance positive or negative? If the positive direction is up the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the automobile's acceleration positive or negative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A. Positive because the car maybe going down but its velocity is still increasing. The air resistance is neg. because it is pushing against the car. B. Then the car would be neg. and the air resistance would be positive. "