Phy 231
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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: ->->->->->->->->->->->-> :
‘dt is 45 cm, v0 is 10 cm/s, vf is 20 cm/s.
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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: ->->->->->->->->->->->-> :
a is 10 cm/s^2, t is 3 sec, and vf is 50 cm/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: ->->->->->->->->->->->-> :
d is 30 cm, a is 20 cm/s^2, v0 is 0 cm/s.
I wouldn't use d as a variable, especially in a calculus-based course.
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
Yes, in the first problem it can be determined. vAve may be found because the vf and v0 are both given. (20 cm/s + 10 cm/s) /2 = 15 cm/s. Therefore, the vAve of the first problem is 15 cm/s.In the second problem, the missing variable is v0 in order to find vAve. I would use the acceleration equation: a = (vf-v0)/’dt. With the given information, the equation becomes 10cm/s^2 = (50 – v0)/3. V0 then becomes 20 cm/s. vAve in this case is (50 cm/s + 20 cm/s) / 2 = 35 cm/s. In the third problem there is not enough information to figure out the vAve.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
dv can be found in the first and second problem but not in the third. In the first problem, dv is vf-v0, which were both given. 20 cm/s – 10 cm/s = 10 cm/s which is dv. In the second problem, vf is 50 cm/s and v0 was found to be 20 cm/s. From there 50 cm/s – 20 cm/s = 30 cm/s. dv is then 30 cm/s in the second problem.
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20 mins
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&#Your work looks good. See my notes. Let me know if you have any questions. &#