cq_1_021

phy 201

Your 'cq_1_02.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Seed Question: Asst 2 Question 1

The problem:

A ball starts with velocity 4 cm/sec and ends with a velocity of 10 cm/sec.

• What is your best guess about the ball's average velocity?

answer/question/discussion: ->->->->->->->->->->->-> :

My best guess about the ball’s average velocity is 6 cm/sec. I derived this by subtracting the end velocity from the start velocity and dividing by 2.

(10cm/sec – 4 cm/sec) / 2

• Without further information, why is this just a guess?

answer/question/discussion: ->->->->->->->->->->->-> :

This is a guess because there are more times that could be taken to give a more accurate average. Two times taken, one at the beginning and end of an interval do not reflect the points between that may increase or decrease in velocity.

• If it takes 3 seconds to get from the first velocity to the second, then what is your best guess about how far it traveled during that time?

answer/question/discussion: ->->->->->->->->->->->-> :

If it takes 3 seconds to get from the first velocity to the second, my best guess about how far it traveled during this time is 18 cm/sec/sec.

We are trying to find the average change in displacement

`d = `v * `t = 6 cm/sec * 3 sec = 18 cm/sec/sec

• At what average rate did its velocity change with respect to clock time during this interval?

answer/question/discussion: ->->->->->->->->->->->-> :

The average rate of velocity change with respect to clock time during this interval is 6 cm.

Vave = Vf – Vo/ `t = (18 cm/s – 0 cm/s)/3 sec = 6 cm

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1 hour

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You are working from the definitions, which is good. You're close, but you're not quite applying them correctly.

&#Please compare your solutions with the expanded discussion at the link

Solution

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