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:
<em>(-6, 0) and (0, 1.5)</em>
<em></em>
Step-by-step explanation:
The equation of the line in pint slope form is expressed as;
y-y0= m(x-x0)
m is the slope
(x0, y0) is the point on the line
Given
m = 1/4
(x0, y0) = (6,3)
Substitute into the formula;
y - 3 = 1/4(x-6)
4(y-3) = x - 6
4y - 12 = x-6
4y - x = -6+12
4y - x = 6
x = 4y - 6
To get the points to plot, we will find the x and y-intercept of the resulting expression.
For the x-intercept,
at y = 0
x = 4(0) - 6
x = -6
Hence the x-intercept is at (-6, 0)
For the y-intercept,
at x = 0
0 = 4y - 6
4y = 6
y = 6/4
y = 3/2
y = 1.5
Hence the y-intercept is at (0, 1.5)
<em>Hence the required points to plot to get the required line are (-6, 0) and (0, 1.5)</em>
<em></em>
Here is the set up:
(3/4)[(11/8) + (11/9) + (7/8)]
Do the math to find your answer.
Answers:
- interest = $75
- balance at maturity = $3075
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Explanation:
The simple interest formula is
i = p*r*t
where in this case,
- p = 3000 = principal (amount deposited)
- r = 0.10 = annual interest rate in decimal form
- t = 3/12 = 0.25 = number of years
So,
i = p*r*t
i = 3000*0.10*0.25
i = 75 is the amount of interest earned
This adds onto the initial deposit to get the final balance when the CD matures (ie when you're able to withdraw the money without penalties)
The balance at maturity is p+i = 3000+75 = 3075 dollars
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In short, you deposit $3000 into the CD and have to wait 3 months for the amount to update to $3075.
2x=y
two times the number of minutes = the amount of pages
