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:
2a^2b(5a-3b)
If not then the other way is
5a-3b
Step-by-step explanation:
The GCF of 10a^3b and -6a^2b^2 is 2a^2b
So divide by that you will get
2a^2b(5a-3b)
Given:
Compound shape
To find:
The area of the compound shape.
Solution:
The compound shape is splitted into two parallelograms.
<u>Bottom parallelogram:</u>
Base = 7.5 cm
Height = 5 cm
Area of the parallelogram = base × height
= 7.5 × 5
= 37.5 cm²
The area of the Bottom parallelogram 37.5 cm².
<u>Top parallelogram:</u>
Base = 7.5 cm
Height = 4.5 cm
Area of the parallelogram = base × height
= 7.5 × 4.5
= 33.75 cm²
The area of the top parallelogram 33.75 cm².
Compound shape = 37.5 + 33.75
= 71.25 cm²
The area of the compound shape is 71.25 cm².
Answer: 3.675 seconds
Step-by-step explanation:
Hi, when the object hits the ground, h=0:
h=−16t^2+48.6t+37.5
0=−16t^2+48.6t+37.5
We have to apply the quadratic formula:
For: ax2+ bx + c
x =[ -b ± √b²-4ac] /2a
Replacing with the values given:
a=-16 ; b=48.6; c=37.5
x =[ -(48.6) ± √(-48.6)²-4(-16)37.5] /2(-16)
x = [ -48.6 ± √ 4,761.96] /-32
x = [ -48.6 ± 69] /-32
Positive:
x = [ -48.6 + 69] /-32 = -0.6375
Negative:
x = [ -48.6 - 69] /-32 = 3.675 seconds (seconds can't be negative)
Feel free to ask for more if needed or if you did not understand something.
Answer:
Angle 2 is equal to angle 8.
Step-by-step explanation:
