Question

A boy moves from point P (1, 2) to point P (3, 4) in X-Y plane in t seconds. Find the magnitude of the displacement vector of the boy in seconds
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Answer 1

Given   : the boy moves from point P (1,2) to P (3,4)

TO FIND: magnitude of displacement vector

from the given condition we have,

$\begin{array}{l}\hat{r}_1\ =\ 1\hat{i}\ +2\hat{j}\end{array}$

$\begin{array}{l}\hat{r}_2\ =\ 3\hat{i}\ +4\hat{j}\end{array}$

displacement vector is given by $\Delta r$

$\Delta r\ =\hat{r}_2\ -\ \hat{r}_1$

$\Delta r\ =\ \left(3\hat{i}\ +4\hat{j}\right)-\ \left(1\hat{i}\ +2\hat{j}\right)$

$\Delta r\ =\ \left(3-1\right)\hat{i}+\left(4-2\right)\hat{j}$

$\Delta r\ =\ 2\hat{i}\ +\ 2\hat{j}$

THE MAGNITUDE OF DISPLACEMENT VECTOR IS GIVEN BY $\left|\Delta r\right|$

$\left|\Delta r\right|\ =\ \sqrt{2^2+2^2}$

$\left|\Delta r\right|\ =\ \sqrt{8}$

Answer 2

$2\sqrt{2}u$ is the magnitude of displacement for the boy.

Explanation:

Given:

Initial position of boy $=P(1,2)=i+2j$

Final position of boy $=Q(3,4)=3i+4j$

To Determine: Magnitude of displacement

Calculation:

From the figure below, the length of the line $PQ$ will give the magnitude of displacement covered by the boy.

$OP = 2 u\\OQ=2u$

So, from Pythagoras theorem,

"The sum of squares of base and perpendicular sides of a right-angled triangle equals to the square of hypotenuse of the triangle".

$PQ=\sqrt{OP^{2}+OQ^{2} }\\PQ=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}u$

Therefore, displacement of boy $=PQ=2\sqrt{2}u$.