Respuesta :
Answer:
a) [tex][0,10]\cup[62,183][/tex], b) [tex][10, 62]\cup[183,+\infty][/tex], c) 3, d) Third-Order Polynomial, negative leading coefficient, e) [tex]t = 183\,s[/tex], f) [tex]2.81g[/tex]
Step-by-step explanation:
a) The intervals over which the acceleration is increasing are:
[tex][0,10]\cup[62,183][/tex]
b) The intervals over which the acceleration is decreasing are:
[tex][10, 62]\cup[183,+\infty][/tex]
c) There are three turning points.
d) The amount of turning points indicates the minimum degree of a polynomial function. The minimum model is a third-order polynomial. The leading coefficient must be negative, since acceleration decreases and turns negative as time goes by and higher degree means a major growth.
e) The greatest acceleration occurs at [tex]t = 183\,s[/tex].
f) The maximum acceleration is [tex]2.81g[/tex]
We want to analyze the given graph of acceleration vs time to answer different things.
A) (0, 10) U (62, 183)
B) (10, 62) U (183, 200)
C) 3
D) A polynomial of degree 3.
E) t = 183s
F) a = 2.81 G-forces.
A) The acceleration is increasing if, reading from left to right, the graph goes up.
By that criterion, the intervals in which the acceleration increases are:
(0, 10) U (62, 183)
B) Now we need to see when the function decreases, this time we need to see that, when reading from left to right, the graph goes downwards.
(10, 62) U (183, 200)
C) We can see that there are 3 turning points, at:
t = 10
t = 62
t = 183
D) There are 3 turning points, so the polynomial of the smallest degree that can represent this is a polynomial of degree 3.
E) We just need to see which is the value of the time when the acceleration is on the highest point of the graph.
We can see that the value of time at that point is:
t = 183 seconds
F) Similar to above, this time we must see the value of the acceleration when it is in the highest point, we can see that the maximum acceleration is:
a = 2.81 G-forces.
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