Particle $A$ is initially at the point with position vector \(\begin{pmatrix} 3 \ -2 \end{pmatrix

Question

Particle $A$ is initially at the point with position vector $\begin{pmatrix} 3 \ -2 \end{pmatrix}$ and moves at a constant speed of $26$ m / s in the direction of the vector $\begin{pmatrix} 12 \ 5 \end{pmatrix}$. Find the position vector of $A$ after $t$ seconds. At the same time when particle $A$ starts moving, particle $B$ starts moving from the point with position vector $\begin{pmatrix} 67 \ -18 \end{pmatrix}$. It travels with speed $20$ m / s at an angle of $\alpha$ above the positive $x$ -axis, where $\tan \alpha = \frac{3}{4}$. Find the position vector of $B$ after $t$ seconds. Hence find the time at which $A$ and $B$ meet, and the position where this occurs.

Accepted Answer

(a) $\begin{pmatrix} 3 + 24t \ -2 + 10t \end{pmatrix}$ (b) $\begin{pmatrix} 67 + 16t \ -18 + 12t \end{pmatrix}$ (c) $t = 8$, $\begin{pmatrix} 195 \ 78 \end{pmatrix}$

Vectors - Vector Kinematics

Vectors - Vector Kinematics