Subscribe Us

Header Ads

Vector (Velocity/ Position/ Speed)

$\def\cvec#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}\def\ABvec#1{\overrightarrow{#1}}$

(CIE 0606/2021/m/12/Q5)
In this question all lengths are in kilometres and time is in hours. Boat A sails, with constant velocity, from a point $O$ with position vector $\cvec{0}{0}$. After 3 hours $A$ is at the

point with position vector $\cvec{-12}{9.}$

(a) Find the position vector, $\ABvec{OP},$ of $A$ at time $t$. [1]

At the same time as $A$ sails from $O$, boat $B$ sails from a point with position vector $\cvec{12}{6},$ with constant velocity $\cvec{-5}{8}$.

(b) Find the position vector, $\ABvec{OQ},$ of $B$ at time $t.$ [1] 

(c) Show that at time $t, |\ABvec{PQ}|^2=26t^2+36t+180.$  [3] 

(d) Hence show that $A$ and $B$ do not collide. [2]

*****\\

*********math solution*************
(a) Let $v=\cvec{v_1}{v_2}$ be velocity vector. 
After $t$ hours, the position vector of $P$ is
$\ABvec{OP}=\cvec 00+t\cvec{v_1}{v_2}$. 
At $t=3$

$\begin{aligned}&\text { } \left(\begin{array}{rl}-12 \\9\end{array}\right)=\left(\begin{array}{l}0 \\0\end{array}\right)+3\left(\begin{array}{l}v_{1} \\v_{2}\end{array}\right)=\left(\begin{array}{l}3 v_{1} \\3 v_{2}\end{array}\right) \\&\therefore \quad v_{1}=-4, \quad v_{2}=3 \\&\therefore \quad \overrightarrow{O P}=\left(\begin{array}{c}0 \\0\end{array}\right)+t\left(\begin{array}{c}-4 \\3\end{array}\right)=\left(\begin{array}{c}-4 t \\3 t\end{array}\right) \\&\text { (b) } \quad \overline{O Q}=\left(\begin{array}{c}12 \\6\end{array}\right)+t\left(\begin{array}{c}-5 \\8\end{array}\right)=\left(\begin{array}{c}12-5 t \\6+8 t\end{array}\right) \\&\text { (c) } \overrightarrow{P Q}=\overline{O Q}-\overline{O P}=\left(\begin{array}{c}12-5 t \\6+8 t\end{array}\right)-\left(\begin{array}{c}-4 t \\3 t\end{array}\right)=\left(\begin{array}{c}12-t \\6+5 t\end{array}\right) \\\end{aligned}$

$\begin{aligned}\therefore|\overrightarrow{P Q}|^{2} &=(12-t)^{2}+(6+5 t)^{2} \\&=\left(144-24 t+t^{2}\right)+\left(36+60 t+25 t^{2}\right) \\&=26 t^{2}+36 t+180 \\\text{(d)    }\qquad&\text { Conrider } 26 t^{2}+36 t+180=0 \\&\text { Discriminant } =b^{2}-4 a c=36^{2}-4 \times 26 \times 180=-17424<0 \\&\text { Thus the equation has no solution. } \\&\therefore |\overrightarrow{P Q}| \neq 0 \text { for any } t . \\&\text { Hence } A \text { and } B \text { do not collide. }\end{aligned} $

**********end math solution********************

Post a Comment

0 Comments