Question

An object goes 10km towards East, 5 km towards West and finally 12km towards North. Find the total distance travelled and the displacement of the object.

Answer 1

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,

$\\ \longrightarrow \quad \sf {Distance = AB + BC + CD }$

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

$\\ \longrightarrow \quad \sf {Distance = (15 + 12) \; km }$

$\\ \longrightarrow \quad \bf \underline {Distance = 27 \; km }$

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.

$\\ \longrightarrow \quad \sf {AC = AB - BC }$

$\\ \longrightarrow \quad \sf {AC = (10 - 5) \; km }$

$\\ \longrightarrow \quad \bf \underline {AC = 5 \; km }$

Now, by using Pythagoras theorem,

$\\ \longrightarrow \quad \bf {AD^2 = AC^2 + CD^2 }$

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

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

$\\ \longrightarrow \quad \sf {(Displacement)^2 = 169 \; km^2 }$

$\\ \longrightarrow \quad \sf {Displacement = \sqrt{169 \; km^2 } }$

$\\ \longrightarrow \quad \bf \underline {Displacement = 13 \; km }$

Therefore, the displacement is 13 km.

Answer 2

The displacement is 13km but above one is perfect explanation.