course Phy 231 6/12 6:15 005. `query 5 Question: `qIntro Prob 6 given init vel, accel, `dt find final vel, dist
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Given Solution: `a**You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. v0 you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the displacement For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = displacement In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just used the equations that we learned in the class notes, is that okay????? I understand how you get the equations
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Given Solution: Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt. Your solution: Dt, v0 and vf are in a line in the middle of the figure. Below that line, v0 and vf are connected to make dv. Then, dv and dt are connected to make a triangle with aAve. Below the second line, v0 and vf are again connected to make a triangle with vAve. Finally, vAve and dt care connected to make a triangle with ds. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first level in the diagram would contain `dt, v0 and vf. From v0 and vf we can easily reason out `dv, so v0 and vf would connect to `dv in the second level. The second level would also contain vAve, also obtained from v0 and vf and therefore connected from vf in the first level to v0 in the first level. The third level would contain an a, which is reasoned out from `dv and `dt and so is connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, which follows from vAve and `dt and is therefore connected to vAve in the second level and `dt in the first level. ** http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/flow_diagrams.htm &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: `qIntro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval? Your solution: First, you can find the displacement using the equation ds = v0(dt)+.5a(dt^2). Then you use the displacement that you just found and use it in the equation ds = (v0+vf/2)*dt. Solve for vf and you have the final velocity. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: `qDescribe the flow diagram we obtain for the situation in which we know v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dt, v0 and vf are in a line in the middle of the figure. Below that line, v0 and vf are connected to make dv. Then, dv and dt are connected to make a triangle with aAve. Below the second line, v0 and vf are again connected to make a triangle with vAve. Finally, vAve and dt care connected to make a triangle with ds. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantities at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok ********************************************* Question: Suppose we have two points on a straight-line graph of velocity vs. clock time. How do we construct a trapezoid to represent the motion on the intervening interval? What aspect of the graph represents the change in velocity for the interval, and why? What aspect of the graph represents the change in clock time for the interval, and why? What aspect of the graph represents the acceleration for the interval, and why? What aspect of the graph represents the displacement for the given interval, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The trapezoid from the points is made by connecting the points with a straight line. Then, straight lines are extended vertically down to the x axis. The points on the axis are then connected. The area inside the lines is the trapezoid. The change in velocity is given by the rise of the graph (the vertical movement). This is because velocity is given on the y axis of the graph. The change in clock time is given by the run (horizontal movement) because clock time is measured on the x axis. The acceleration is the slope of the graph because it is the change in velocity (dy) over the change in time (dx). The displacement is given by the area under the curve. It is equivalent to the average velocity times the change in clock time, which gives a distance when simplified. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Question: `qPrinciples of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: New York to California = 2500 miles (1.609km/1mi) = 4022.5 km convert to km 4022.5km / (10km/hr) = 402.25 hours use the equation vAve=ds/dt, solved for dt confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr. Be sure you understand the units of this calculation. Units should be used at every step of every calculation. The corresponding symbolic solution: vAve = `ds / `dt; we want to find `dt so we solve to get `dt = `ds / vAve. Substituting `ds = 5000 km and vAve = 10 km/hr we have `dt = 5000 km / (10 km/hr) = 500 hr. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result? Your solution: Average heart rate = 75 beats / minute Life expectancy = 75 years 75bpm (60min/1hr)(24hr/1day)(365.25days/1yr)(75yr/1life) = 2958525000 = 2.96*10^9 beats in a lifetime confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion beats, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, but my assumptions were different Self-critique rating #$&* Ok Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors? Your solution: First, find the dot product: -2(2) + 6(-3) = -4-18 = -22 Magnitude of A = sqrt ((-2)^2 +(6^2)) = sqrt(4+36) = sqrt(40) Magnitude of B = sqrt((2^2)+(-3)^2) = sqrt(4+9) = sqrt(13) Scalar product: ABcos(theta) = dot product of A and B Sqrt(40)*sqrt(13)*costheta = -22 Cos(theta) = -22 / sqrt(40*13) Theta = arccos(-22 / sqrt520) = 164.74 degrees = 165 degrees confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. ** Self-critique (if necessary): OK! Self-critique rating #$&* OK Add comments on any surprises or insights you experienced as a result of this assignment. None "