Answer:
34
Step-by-step explanation:
The acceleration of the particle is given by the formula mentioned below:
Differentiate the position vector with respect to t.
![\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bds%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Csqrt%5B%5D%7B%5Cmleft%28t%5E3%2B1%5Cmright%29%7D%20%5C%5C%20%3D-%5Cfrac%7B1%7D%7B2%7D%28t%5E3%2B1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ctimes3t%5E2%20%5C%5C%20%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bt%5E2%7D%7B%5Csqrt%7B%28t%5E3%2B1%29%7D%7D%20%5Cend%7Bgathered%7D)
Differentiate both sides of the obtained equation with respect to t.
![\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2s%28t%29%7D%7Bdx%5E2%7D%3D%5Cfrac%7B3%7D%7B2%7D%28%5Cfrac%7B2t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D%2Bt%5E2%28-%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%29%20%5C%5C%20%3D%5Cfrac%7B3t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D-%5Cfrac%7B9%7D%7B4%7D%5Cfrac%7Bt%5E2%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cend%7Bgathered%7D)
Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
![\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%28t%3D1%29%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B2%7D%7D-%5Cfrac%7B9%7D%7B4%5Ctimes2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%20%3D1.32ft%2Fsec%5E2%20%5Cend%7Bgathered%7D)
The initial position is obtained at t=0. Substitute t=0 in the given position function.
Answer:
graph g(x)=1/4 x^2 - 2
Step-by-step explanation:
You are to replace x with (1/2x) in the expression x^2-2
So you have (1/2x)^2-2
1/4 x^2-2
Graph some points for g(x)=1/4 x^2-2
The vertex is (0,-2) and the parabola is open up.
I would graph 2 more points besides the vertex
x | g(x) ordered pairs to graph
----------- (-1,-1.75) and (0,-2) and (1,-1.75)
-1 -1.75
0 -2
1 -1.75
Answer:
45
Step-by-step explanation:
AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.
now, what is f ?
we know this from the relation of AC to FA.
FA = 12 mm
AC = 12 + 28 = 40 mm
so, going from AC to FA we multiply AC by f so that
AC × f = FA
40 × f = 12
f = 12/40 = 3/10
all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).
the area of a triangle is
baseline × height / 2
aABC = 500
and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.
baseline small = baseline large × f
height small = height large × f
we know that
baseline large × height large / 2 = 500
baseline large × height large = 1000
aAEF = baseline small × height small / 2 =
= baseline large × f × height large × f / 2 =
= baseline large × height large × f² / 2 =
= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =
= 500 × 9/100 = 5 × 9 = 45 mm²
Plug in "8i + 15j + k" for "f", and "2i – 5j + 9k" for "e"
3(8i + 15j + k) - 2(2i – 5j + 9k)
Distribute 3 to (8i + 15j + k) & - 2 to (2i – 5j + 9k)
3(8i + 15j + k) = 24i + 45j + 3k
-2(2i - 5j + 9k) = -4i + 10j - 18k
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Simplify. Combine like terms
24i - 4i + 45j + 10j + 3k - 18k
(24i - 4i) + (45j + 10j) + (3k - 18k)
20i + 55j - 15k is your answer, or (C)
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hope this helps
