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course Phy 121

10/31 615Revision to Seed problem concerning area of graph as requested

cq_1_041

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Phy 121

Your 'cq_1_04.1' 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_04.1_labelMessages **

The problem:

A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock

time is 9 seconds.

Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and

(9 s, 40 cm/s).

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Velocity is vertical axis, time represents horizontal

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Sketch a straight line segment between these points.

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

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What are the rise, run and slope of this segment?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

rise= 40-10 cm/s = 30cm/s

run= 9sec-4sec = 5sec

slope= 30cm/s / 5sec = 6cm/s^2

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What is the area of the graph beneath this segment?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

A= (30cm/s^2 * 5s) / 2 = (150cm/s^2)/2 = 75cm/s^2

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A= [(10cm/s + 40cm/s) / 2] * 5s = 25cm/s * 5s = 125cm

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30 cm/s^2 isn't a quantity associated with this situation.

Your graph consists of a trapezoid with 'graph altitudes' 10 cm/s and 40 cm/s, and

width 5 s.

The area of this trapezoid is equal to that of a rectangle of the same width whose 'graph

altitude' is the average of the two altitudes.

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Good answers on rise, run and slope.

You didn't get the area correctly, so I'm going to ask you to revise that

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