Question

Two particles $P$ and $Q$ are moving with constant velocities of $(\hat{i}+\hat{j})m/s$ and $(-\hat{i}+2\hat{j})m/s$ respectively. At time $t=0,P$ is at origin and $Q$ is at a point with position vector $(2\hat{i}+\hat{j})m$. If the equation of the trajectory of $Q$ as observed by $P$ is $x+2y=n$, then find $n958$.

Answer 1

velocity of point P is

${\overrightarrow{v}}_p$ = $(\hat{i}+\hat{j})$

velocity of Q is

${\overrightarrow{v}}_q=(-\hat{i}+\hat{2}j)$

Displacement of point P is

$x_1\hat{i}+y_1\hat{j}=(\hat{i}+\hat{j})\times t$

displacement of pont Q is

$(x_2-2)\hat{i}+(y_2-1)\hat{j}=(-2\hat{i}+\hat{j})\times t$

Displacement of point Q with respect to P is

$\left(\right(x_2-2)\hat{i}+(y_2-1\left)\hat{j}\right)-(x_1\hat{i}+y_1\hat{j})$

= $(-\hat{i}+2\hat{j})\times t-(\hat{i}+\hat{j})\times t$

$(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}-2\hat{i}-\hat{j}$

= $(-\hat{i}+2\hat{j})\times t-(\hat{i}+\hat{j})\times t$

$(x-2)\hat{i}+(y-1)\hat{j}$

= $(-2\hat{i}t+t\hat{j})\left(\right(x_2-x_1)$ = x and $(y_2-y_1)=y)$

$x$ - 2 = -2t . . . . 1

and y - 1 = t . . . . 2

$x$ - 2 = - 2( y -1 )

$x$ - 2 = -2y + 2

2y + $x$ = 4

comparing with the given equation

n = 4