Question

Find the
• distance
• displacement

[ Remember - CE = ED = 3 m ]

Answer 1

In the question it is asked to find the distance and displacement

1) Distance

So we know that , distance is the length of the path travelled by a body in certain interval of time .

$\begin{gathered} \longmapsto \rm { Distance = AB + BC + CD}\\ \end{gathered}$

Substituting the values

$\begin{gathered} \longmapsto \rm { Distance = 3 + 4 + 6}\\ \end{gathered}$

Performing addition

$\begin{gathered} \longmapsto \rm \bold{ Distance = 13 m}\\ \end{gathered}$

2) Displacement

We know that , Displacement is measured as the shortest distance (straight line) between the initial and final position of a moving body .

• Initial position - A
• Final position - D

In order to calculate the Displacement, we need to find the length of AD. Since AED is a right-angled triangle, so by using pythagoras theorem,

$\begin{gathered} \longmapsto \rm { (AD)^2 = (AE)^2 + (ED)^2}\\ \end{gathered}$

Substituting the values

$\begin{gathered} \longmapsto \rm { (AD)^2 = (4\; m)^2 + (3\; m)^2}\\ \end{gathered}$

$\begin{gathered} \longmapsto \rm { (AD)^2 = 16 \; m^2 + 9\; m^2}\\ \end{gathered}$

Performing addition

$\begin{gathered} \longmapsto \rm { (AD)^2 = 25 \; cm^2}\\ \end{gathered}$

$\begin{gathered} \longmapsto \rm { AD = \sqrt{25 \; m^2}}\\ \end{gathered}$

$\begin{gathered} \longmapsto \rm { AD = \bold{5 \; m}}\\ \end{gathered}$

$\begin{gathered} \longmapsto \rm \bold{ Displacement = 5 m}\\ \end{gathered}$

Answer 2

Answer:

sorry I don't know answer of your question