I need help with my hinework! I already have a D for math 2a+ 13-0+65b
1 answer:
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Answer:
(a) d = 0.012v^2
(b) 4.8 m
(c) 25 m/s
Step-by-step explanation:
(a) Writing an equation to fit a bunch of data is often a matter of trial and error. You can often get an idea of the sort of function that is involved by graphing the data points.*
When we do that, we see the relationship is non-linear with an upward curvature. It might be described by a polynomial or exponential function.
When we try a quadratic equation, we find the fit is really quite good. The graphing calculator that does this quadratic regression analysis tells us the equation is essentially ...
d = 0.012·v^2
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(b) The equation predicts a depth of 4.8 m.
d = 0.012·20^2 = 4.8
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(c) Solving for speed, we get a speed of 25 m/s.
7.5 = 0.012v^2
625 = v^2 . . . . . . divide by .012
25 = v . . . . . . . . . take the square root
_____
* The energy of a falling object is proportional to the square of its speed. If we assume that the depth of penetration into the clay is proportional to the energy absorbed, then it is not surprising to see a quadratic relationship between speed and distance. What is surprising is that this relationship would show up in middle school math problem, not a high-school physics problem.
Essentially, you have to assume the form of the equation, then find the coefficients that make it so. Working with the evenly spaced table values of 5, 10, 15 m/s can simplify the effort.
A graphing calculator or spreadsheet can do the bulk of the work once the data is entered.
Recall your d = rt, distance = rate * time
so, notice, it went 504 against the wind, and the return trip, well, unless the road elongated or shrunk somewheres, the return trip is also 504 miles.
now, if say the plane has a still air speed of "p", and the wind has a speed of "w", when it was going against the wind, it wasn't really going "p" fast, it was going "p - w" fast, because the wind was subtracting speed from it.
now, when it was going with the wind, it wasn't going "p" fast either, it was going faster, at "p + w" fast, because the wind was adding to it. thus
so, notice, it went 504 against the wind, and the return trip, well, unless the road elongated or shrunk somewheres, the return trip is also 504 miles.
now, if say the plane has a still air speed of "p", and the wind has a speed of "w", when it was going against the wind, it wasn't really going "p" fast, it was going "p - w" fast, because the wind was subtracting speed from it.
now, when it was going with the wind, it wasn't going "p" fast either, it was going faster, at "p + w" fast, because the wind was adding to it. thus