Question

A particle first moves 4m due East and then 3m due south . Determine the displacement and distance travelled by particle .​

Answer 1

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✦ Required Answer:

♦️ GiveN:

• The body moves 4 m East and then 3 m west.

♦️ To FinD:

• Find distance....?
• Find displacement.....?

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✦ How to Solve?

Before solving this question, we need to know the actual definitions of distance and displacement, especially their difference. The $\large{\rm{\purple{Distance}}}$ is the actual path covered by the body. The total length of the path travelled, whereas $\large{\rm{\purple{Displacement}}}$ is the shortest path between the initial and final point. It doesn't depend the path covered by body, only we should consider the straight line joining the initial and final point to be the displacement.

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✦ Solution:

Refer to the attachment...

Let the body started its journey from A, and travels 4 m east to reach B. Then it travels 3 m south to reach point C. So, the body travels from A to B, and then B to C. So, total distance travelled AB + BC.

$\large{ \rm{ \longrightarrow \: AB + BC= 4m + 3m }} \\ \\ \large{ \rm{ \longrightarrow \: Total \: distance \: = \boxed{ \rm{ \green{7m}}}}}$

Now, we have initial point A and final point C, so we know that displacement is the shortest path between initial and final point. So, displacement will be AC.

By using pythagoras theoram,

$\large{ \rm{ \longrightarrow \: {AC}^{2} = {AB}^{2} + {BC}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: AC = \sqrt{ {AB}^{2} + {BC}^{2} }}} \\ \\ \large{ \rm{ \longrightarrow \: AC = \sqrt{ {4}^{2} + {3}^{2} } \: m}} \\ \\ \large{ \rm{ \longrightarrow \: AC = \sqrt{16 + 9} \: m}} \\ \\ \large{ \rm{ \longrightarrow \: AC = \sqrt{25} \: m}} \\ \\ \large{ \rm{ \longrightarrow \: Displacement = \: \boxed{ \rm{ \green{5 \: m}}}}}$

$\large{ \therefore{ \underline{ \underline{ \purple{ \rm{Hence, \: solved \: \dag}}}}}}$

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