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

Answer:

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

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

The shortest distance is thus given by

$r^{2}=(4-2)^{2}+(4-6)^{2}\\\\r=2\sqrt{2}$