Hi everyone I am in need of some guidance solving this parametric equation question and was wondering if you guys could give me some pointers and to see if I am doing this correctly.
Here I have two functions:
$$y=1600-710t+90t^2$$ $$and$$ $$x=-100+250t-45t^2$$
my goal is to compute the values of its position, direction, velocity, and acceleration at $$t=0.8 s$$
so far from what I (hopefully) got correct is in my work below.
In calculating the position of the object moving along this path I've done the following:
inputting the value at $t=0.8s$ into the functions of $x$ and $y$
$$y=1600-710(0.8)-90(0.8)^2 $$ $$y=974.4\;{m}$$ $$x=-100+250(0.8)-45(0.8)^2$$ $$x=71.2\;{m}$$ Therefore: $$(x,y)=(71.2\;{m},974.4\;{m})$$
For Velocity, the following is:
taking the derivative of each function at $t=0.8\;{s}$ giving:
$$y'=-710+180t$$ $$v_y'=-566\;{m/s}$$ $$x'=250-90\;{m/s}$$ $$v'_x=175\;{m/s}$$
$$\vec{v}=\sqrt{{175}^2+{(-566)}^2} = 593.32\;{m/s}$$ $$\theta = Tan^{-1}{(\frac {-566}{175})} = -72.54^\circ$$
Therefore the velocity of the moving object at $t=0.8\;{s}$ is $593.32\;{m/s}$ at $\theta$ = $-72.54^\circ$
For Acceleration, taking the second-derivative:
$$a_x=-90\;{m/s^2}$$
$$a_y=180\;{m/s^2}$$
$$\vec{a}=\sqrt{{-90}^2+{180}^2}=201.2\;{m/s^2}$$ $$\theta=Tan^{-1}(\frac{180\;{m/s^2}}{-90m\;{m/s^2}}) = -63.43^\circ $$
Therefore the object is accelerating at $201.2\;{m/s^2}$ at $\theta =-63.43^\circ$