course Phy 121 006. `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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance. If the positive direction is down the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the air resistance positive or negative? If the positive direction is up the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the automobile's acceleration positive or negative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity is positive, acceleration is positive, and air resistance is negative because the direction down the hill is positive. And the exact opposite for going up the hill. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you are riding in the car your perception is that the forward direction is the one in which you are moving. You can stick your hand out the window and feel that the air resistance is in the 'backward' direction. Since the automobile is speeding up its acceleration is in its direction of motion. The velocity is down the hill. Thus the direction of the velocity and acceleration are both down the hill, and the direction of the air resistance is up the hill. Therefore • If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. • If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "