Question

Two particles A and B have velocities 3î and 4î + ſ respectively. (i) Find rbiat), the position of B relative to A for all t given that rB|A(t = 0) = –2î – 6ſ. 15 Marks (ii) Show that A and B are closest together when t = 4 and hence find the shortest distance between them.

Answer 1

Given that

`\overrightarrow{v_(a)}=3\widehat{i}\\\\\overrightarrow{v_(b)}=4\widehat{i}+\widehat{j}\\\\\therefore \overrightarrow{v_(ba)}=4\widehat{i}+\widehat{j}-3\widehat{i}\\\\\overrightarrow{v_(ba)}=\widehat{i}+\widehat{j}\\\\\frac{\overrightarrow{dr_(ba)}}{dt}=\widehat{i}+\widehat{j}\\\\\therefore \overrightarrow{r_(ba)}=\int (\widehat{i}+\widehat{j})dt\\\\\therefore \overrightarrow{r_(ba)}=t\widehat{i}+t\widehat{j}+\overrightarrow{r_(o)}`

Now it is given that

`\overrightarrow{r_(o)}=-2\widehat{i}-6\widehat{j}`

`\therefore \overrightarrow{r_(ba)}=(t-2)\widehat{i}+(t-6)\widehat{j}`

Now the distance can be calculates as

`r=\sqrt{x^(2)+y^(2)}\\\\r^(2)=(t-2)^(2)+(t-6)^(2)\\\\`

Differentiating with respect to 't' and equating to zero for minimizing the function we get

`r=\sqrt{x^(2)+y^(2)}\\\\r^(2)=(t-2)^(2)+(t-6)^(2)\\\\2r(dr)/(dt)=2(t-2)+2(t-6)\\\\2r(dr)/(dt)=4t-16\\\\\therefore (dr)/(dt)=0\\\\\Rightarrow 4t-16=0\\\\\therefore t=4`

The shortest distance is thus given by

`r^(2)=(4-2)^(2)+(4-6)^(2)\\\\r=2√(2)`