The equation is y=2.5x.
Let the y axes increase by 2.5 and the x axes by ones
X=1 so it is 1×1×1=1 so.... it is 19+1
The position function of a particle is given by:
![X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t](https://tex.z-dn.net/?f=X%5Cmleft%28t%5Cmright%29%3D%5Cfrac%7B2%7D%7B3%7Dt%5E3-%5Cfrac%7B9%7D%7B2%7Dt%5E2-18t)
The velocity function is the derivative of the position:
![\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%28t%29%3D%5Cfrac%7B2%7D%7B3%7D%283t%5E2%29-%5Cfrac%7B9%7D%7B2%7D%282t%29-18%20%5C%5C%20%5Ctext%7BSimplifying%3A%7D%20%5C%5C%20V%28t%29%3D2t%5E2-9t-18%20%5Cend%7Bgathered%7D)
The particle will be at rest when the velocity is 0, thus we solve the equation:
![2t^2-9t-18=0](https://tex.z-dn.net/?f=2t%5E2-9t-18%3D0)
The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:
![\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%2B15%7D%7B4%7D%3D6%20%5C%5C%20t%3D%5Cfrac%7B9-15%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
(I know there isnt any real problem here to solve, but heres a tip on how to solve greatest to least with decimal problems)
1. Just because a number looks big, doesn't mean it is big
Example: 1.00000000000001 < 1.1
just look at the numbers ^ and dont just hastily read it over assuming that 1.1 is smaller because it "has less digits"
2. Negative numbers are... opposite. and they are less than positive numbers
-3.4 > -4.1
Why is this? Well, if you look on a line, with the point 0 in the middle, you can see that -3.4 is not as far away from 0 as -4.1 is. So the idea is to apply opposite logic for negative numbers
I hope these tips helped!! :D