Question

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Answer 1

Question;

When you travel 40 km East and then 10km towards WEST , your net displacement is? by

Answer ;

Distance = 27 km

Displacement = 13 km

Explanation:

As per the provided information in the given question, we have (refer to the attachment for better understanding) :

An object goes 10km towards East (AB).

Then, 5 km towards west. (BC)

After that, it goes 12 km towards north. (CD)

★ Calculating distance travelled :

Distance means total length of the path covered by the body in the entire journey. So,

\begin{gathered} \\ \longrightarrow \quad \sf {Distance = AB + BC + CD } \\\end{gathered}

⟶Distance=AB+BC+CD

\begin{gathered} \\ \longrightarrow \quad \sf {Distance = (10 + 5 + 12) \; km } \\\end{gathered}

⟶Distance=(10+5+12)km

\begin{gathered} \\ \longrightarrow \quad \sf {Distance = (15 + 12) \; km } \\\end{gathered}

⟶Distance=(15+12)km

\begin{gathered} \\ \longrightarrow \quad \bf \underline {Distance = 27 \; km } \\\end{gathered}

⟶

Distance=27km

Therefore, distance travelled is 27 km.

★ Calculating displacement :

Displacement is the shortest distance from initial position to final position.

Here, A is its initial position and D is its final position. Shortest distance from A to D is a straight line that is AD. We'll be using Pythagoras theorem to find the length of ADm Before that, we need to find the length of AC.

\begin{gathered} \\ \longrightarrow \quad \sf {AC = AB - BC } \\\end{gathered}

⟶AC=AB−BC

\begin{gathered} \\ \longrightarrow \quad \sf {AC = (10 - 5) \; km } \\\end{gathered}

⟶AC=(10−5)km

\begin{gathered} \\ \longrightarrow \quad \bf \underline {AC = 5 \; km } \\\end{gathered}

⟶

AC=5km

Now, by using Pythagoras theorem,

\begin{gathered} \\ \longrightarrow \quad \bf {AD^2 = AC^2 + CD^2 } \\\end{gathered}

⟶AD

2

=AC

2

+CD

2

\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = (5 \; km)^2 + (12 \; km)^2 } \\\end{gathered}

⟶(Displacement)

2

=(5km)

2

+(12km)

2

\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = 25 \; km^2 + 144 \; km^2 } \\\end{gathered}

⟶(Displacement)

2

=25km

2

+144km

2

\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = 169 \; km^2 } \\\end{gathered}

⟶(Displacement)

2

=169km

2

\begin{gathered} \\ \longrightarrow \quad \sf {Displacement = \sqrt{169 \; km^2 } } \\\end{gathered}

⟶Displacement=

169km

2

\begin{gathered} \\ \longrightarrow \quad \bf \underline {Displacement = 13 \; km } \\\end{gathered}

⟶

Displacement=13km

Therefore, the displacement is 13 km.

is your answer :-)

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