cq_1_041

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Phy 231

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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).

Sketch a straight line segment between these points.

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

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

Rise = `dv =(40cm/s-10cm/s) =30cm/s

Run = `dt =(9s-4s) =5s

Slope =rise/run =(30cm/s)/5s =6cm/s^2 (aAve)

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What is the area of the graph beneath this segment?

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

area= base*average height

average height = (height 1 + height 2)/2 =(vInitial+vFinal)/2 = (10cm/s+40cm/s)/2 =(50cm/s)/2 =25cm/s

base= run= 5s

area= (5s)(25cm/s)= 125cm

Since the area ends up being in units of distance, this tells us that the area represents total displacement of the ball.

So the ball is displaced 125 cm.

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&#Very good responses. Let me know if you have questions. &#