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PHY 201
Your 'cq_1_06.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_06.1_labelMessages **
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each.
• A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion: ->->->->->->->->->->->-> :
v_0 = 10 cm/s, v_f = 20 cm/s, ‘ds = 45 cm
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• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion: ->->->->->->->->->->->-> :
v_f = 50 cm/s, a = 10 cm/s^3, ‘dt = 3 s
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• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion: ->->->->->->->->->->->-> :
‘ds = 30 cm, v_0 = 0 cm/s, a = 20 cm/s^2
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
1. Yes, (v_f + v_0)/2 = (20 + 10)/2 = 5 cm/s
2. Not directly, v_0 = v_f - a*’dt = 50 - 10*3 = 20 cm/s then (20+50)/2 = 35 cm/s. Thinking about it a little more, you could assume that you could have v_ave = v_f - .5*a*dt = 50-.5*10*3 = 50 - 15 = 35 cm/s, since we know acceleration is constant.
3. I don’t think you can directly do it. You can use v_f^2 = v_0^2 +2*a*’ds = 0 + 2*20(cm/s^2)*30(cm) = 1200 (cm^2/s^2) so v_f = sqrt(1200 (cm^2/s^2)) = 34.6 m/s. Then find the average v = (v_f + v_0)/2 = 17.3 m/s. Which this is half of v_f but with the squared exponents, you can’t multiply by .5.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
1. Yes, v_f - v_0 = 20 m/s - 10 m/s = 10 m/s
2. Yes, a*’dt will give you the change in velocity so 10 m/s^2 * 3 s = 30 m/s
3. Yes, since v_0 is 0 we know that v_f will also be ‘dv. Work the v_f^2 equation as above and find v_f = 34.6 m/s which we know is also ‘dv
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&#Good responses. Let me know if you have questions. &#