Position vector

The position vector of a point mass moving in a plane can be expressed in the established reference frame by the relation:
r(t) = (1 + 5t + 2t² ; 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 = 3 s?
b) The magnitude of the velocity of the point mass at time t = 1 s
c) The magnitude of the acceleration of the point mass at time t = 4 s

Final Answer:

x =  34
y =  10
v =  9.4868 m/s
a =  4 m/s²

Step-by-step explanation:

$$\begin{gathered} t_1=3\text{s} \\ r(t)=(1+5t+2t^2;3t+1) \\ x=1+5\cdot t_1+2\cdot t_1^2=1+5\cdot 3+2\cdot 3^2=34 \end{gathered}$$

$y=3\cdot t_1+1=3\cdot 3+1=10$

$$\begin{gathered} t_2=1\text{s} \\ v(t)=r^{\prime }(t)=(5+4t;3) \\ {v}_0=5+4\cdot t_2=5+4\cdot 1=9 \\ {v}_1=3 \\ v=\sqrt{v_0^2+v_1^2}=\sqrt{9^2+3^2}=3\sqrt{10}=9.4868\text{m/s} \end{gathered}$$

$$\begin{gathered} t_3=4\text{s} \\ a(t)=r^{\prime \prime }(t)=(4;0) \\ a=\sqrt{4^2+0^2}=4{\text{m/s}}^2 \end{gathered}$$

Try another example