Vector (Kinematic/ Velocity/ Speed)

CIE 0606/2020/W/11/No 6 $\def\cvec#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}$

A particle P is initially at the point with position vector $\cvec{30}{10}$  and moves with a constant speed of 10ms$^{-1}$ in the same direction as $\cvec{-4}{3}.$

(a) Find the position vector of $P$ after $t$ s. [3]

As $P$ starts moving, a particle $Q$ starts to move such that its position vector after $t$ s is given by $\cvec{-80}{90}+t\cvec{5}{12}.$

(b) Write down the speed of $Q.$ [1]

(c) Find the exact distance between $P$ and $Q$ when $t= 10$ , giving your answer in its simplest surd form. [3]

*********math solution*************

(a) Let $v=\cvec{-4}{3}k.$  Then speed =$|v|=10=\sqrt{(-4k)^2+(3k)^2}=5k.$ Thus $k=2.$

Hence the velocity=$2\cvec{-4}{3}=\cvec{-8}{6}.$

$\therefore$ position vector of $P$ after $t$ s$=\cvec{30}{10}+\cvec{-8}{6}t$.

(b) Velocity of $Q=\cvec{5}{12}.$

Thus speed of $Q=\sqrt{5^2+12^2}=13$ m/s.

(c) When $  t=10,$ position vector of $P=\cvec{30}{10}+\cvec{-8}{6}\times 10=\cvec{-50}{70}.$

When $t=10,$ position vector of $Q=\cvec{-80}{90}+\cvec{5}{12}\times 10=\cvec{-30}{210}.$

$\begin{array}{rcll}\therefore PQ&=&\sqrt{(-30+50)^2+(210-70)^2}\\&=&\sqrt{20^2+140^2}\\ &=& 100\sqrt 2\end{array}$

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