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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Introductory Problem 1.3
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In introductory problem 1.3, I am a little confused as to how we solve it. I feel as though there is a simple answer, but I just can't get my mind to wrap around it.
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We are given the question: How far will an object move in 4.4361 seconds if it is moving at a constant speed of .00039 meters/second?
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Would you not just set it up as .00039m / 1s * 4.4361s and then cross multiply and divide, or am I missing an important step?
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The idea of 'cross multiplication' is a common point of confusion and I recommend that you avoid using it wherever possible. All that is involved here is multiplication of fractions, which has nothing to do with cross multiplication.
In this case your calculation is
.00039m / (1s) * 4.4361s
just as you say. There is no 'cross multiplication' involved. You simply have a fraction .00039 m / (1 s) which you multiply by 4.4361 s.
The numerator of the result is
.0039 m * 4.4361 s
and the denominator is 1 s.
The result is therefore
(.0039 m * 4.4361 s) / (1 s)
= .02 m * s / (1 s)
= (.02 / 1) * (m * s / s)
= .02 m
where the .02 is a rough approximation of .0039 * 4.4361.
This is rigorously correct, though worth understanding, but it might be a little confusing.
To put it more simply, but not quite as rigorously:
.0039 m * 4.4361 s = .02 m * s, approximately.
.02 m * s / (1 s) = .02 m
since s / s = 1.
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