b. During a Saturday morning, Mary decides to go on a hike at the Rock Creek Park. Mary is trying to calculate how many miles she walks per hour to try and increase her speed. Table 1.1 gives her miles hiked, y, at time (hours), t. What is Mary's velocity at exactly 1 hour?
c.
Table 1.1 Mary's Miles per Hour for Hiking
t (hours)
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0
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1
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2
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3
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4
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5
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y = m (t) (miles/hour)
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0
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4
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7.5
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10
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12
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15
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d.
Figure 1.1 Graph of Mary's Velocity (miles/hour) over a period of time
e. Now, I will be calculating the slope (ARC) of three secant lines based on x = 1 (Point P = t =1).
1. The first secant line will be from 1 to 2 hours; (1, 4) (2, 7.5).
Calculation: (7.5 - 4)/(2 - 1) = 3.5 miles/hour
2. The second secant line will be from 2 to 3 hours; (2, 7.5) (3, 10).
Calculation: (10 - 7.5)/(3 - 2) = 2.5 miles/hour
3. The third secant line will be from 3 to 4 hours; (3, 10) (4, 12).
Calculation: (12 - 10)/ (4 - 3) = 2 miles/hour
Based off of these calculations it seems that Mary's velocity is decreasing as the number of hours go by. This can be accepted because the average person's velocity tends to decrease as the more hours go by because of tiredness and of energy loss.
Next, I will do 3.5 + 2.5 + 2 / 3 = 2.67 miles/hour. This calculation shows that Mary's slope (ARC) for hiking is approximately 2.67 miles/hour.
f. Tangent line that passes though the point P = 1
g. The second point on this graph represented as Q will be used to find the IRC for PQ, P as defined in part
f. Point P is (1, 3.85) and Point Q is (3.2, 10.5) where the x values represent time (hours) and the y values represent distance (miles).
Slope (IRC) Calculation: (10.5 - 3.85)/(3.2 - 1) = 1.24 miles/hour.
In terms of this application, I was able to calculate the IRC of Mary's velocity which was about 1.24 miles/hour. It makes sense that the IRC is smaller than the ARC of 2.67 miles/hour because I am zooming in on the graphed line, creating more of an accurate representation of Mary's hiking velocity.
h. In this experiment, I was identifying Mary's velocity while she was hiking for a period of time. The tangent line passed through Point P (1, 3.85) and Point Q (3.2, 10.5), which allowed me to calculate the IRC for this experiment which was 1.24 miles/hour. Thus, the line would increase at a rate of 1.24 miles/hour. Based on this information, Mary needs to train more in order to increase her velocity for hiking.
I liked you application of speed! It's true that the faster you move the more distance you cover, yet there are certain instances where you are going faster or slower than you previously were so this is a great way to find out speed at an exact time!
ReplyDeleteGreat application of the the IRC.
ReplyDeletejessica,
ReplyDeletereally great application! i love hiking, so i loved your back story. :) your graphs and tables look good and are well organized. and your explanations of your results are clear and thoughtful. there is a slight issue with your calculations in the beginning with the secant lines. you calculated secant lines from 1 to 2, 2, to 3 and 3 to 4. when you should have calculated all of the secant lines as being connected to the point t = 1, since you were wanting to find the IRC at exactly one hour. so your secant line calculations should have been from 1 to 2, 1 to 3, and 1 to 4. it looks like it didn't change the results of your experiment too much, but just be aware of this for future reference, as the secant line calculations are supposed to should lines that are calculated closer and closer to the point that you want.
other than that, you did a wonderful job!
professor little