Position vector of a point mass

The position vector of a point mass moving in a plane can be expressed in the established reference frame by the relation:
r(t) = (6t²+ 4t ; 3t + 1)
where t is time in seconds and the vector coordinates are in meters.
Calculate:
a) What is the position of the point mass at time t = 2 s?
b) The magnitude of the velocity of the point mass at time t = 3 s
c) The magnitude of the acceleration of the point mass at time t = 5 s

Final Answer:

x =  32
y =  7
v =  40.1123 m/s
a =  12 m/s²

Step-by-step explanation:

$$\begin{gathered} t_1=2\text{s} \\ r(t)=(6t^2+4t;3t+1) \\ x=6\cdot t_1^2+4\cdot t_1=6\cdot 2^2+4\cdot 2=32 \end{gathered}$$

$y=3\cdot t_1+1=3\cdot 2+1=7$

$$\begin{gathered} t_2=3\text{s} \\ v(t)=r^{\prime }(t)=(12\cdot t+4;3) \\ {v}_0=12\cdot t_2+4=12\cdot 3+4=40 \\ {v}_1=3 \\ v=\sqrt{v_0^2+v_1^2}=\sqrt{40^2+3^2}=\sqrt{1609}=40.1123\text{m/s} \end{gathered}$$

$$\begin{gathered} t_3=5\text{s} \\ a(t)=r^{\prime \prime }(t)=(12;0) \\ a=\sqrt{12^2+0^2}=12{\text{m/s}}^2 \end{gathered}$$

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