The required graph is attached.
Line segment XY is tangent to circle Z at point U. Circle Z is shown. Line segment U V is a secant. Line segment X Y is a tangen
2 answers:
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Answer:
(2,8)
x = 2
y = 8
Step-by-step explanation:
<u>The steps are shown in the picture</u>
Answer:
a)
b)
c)
When the acceleration is 0 we have:
And if we replace t=2 in the velocity function we got:
Step-by-step explanation:
For this case we have defined the following function for the position of the particle:
Part a
From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:
Part b
For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:
We can factorize this function as
So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is we need to analyze the following intervals:
For this case if we select two values let's say 0.25 and 0.5 we see that
And we see that for we have that f(b)>f(a) so then the function is decreasing on this case.
We have a minimum at t=2 since at this value w ehave the vertex of the parabola :
And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75
v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2
We see that the function would be increasing.
For this interval we will see that for any two points a,b with we have
for example let's say a=3 and b =4
The particle is moving to the right then the velocity is positive so then the answer for this case is:
Part c
When the acceleration is 0 we have:
And if we replace t=2 in the velocity function we got: