Question

3. A man goes to his office from his house 30 km away in the morning and returns back to his home in the evening. calculate the distance and displacement covered by the man.

Answer 1

Answer:

here is your answer

Explanation:

3 * 2) = (2 * 3) = 6 Km

2 hours going and 3 hours returning.

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Answer 2

Provided that:

A man goes to his office from his house 30 km away in the morning and returns back to his home in the evening.

To calculate:

• His displacement
• The distance covered by him

Solution:

★ His displacement = 0 m

★ The distance covered by him = 60 m

Knowledge required:

→ If initial and final positions are same then the displacement is equal to zero.

Required solution:

~ Firstly let us find out the distance taken by the man!

Explanation: As he go and came back to and from a distance of same 30 kilometres. Henceforth, the total distance must be equal to 60 because

→ Total distance = Distance A + Distance B

→ Total distance = (30 + 30) kilometres

→ Total distance = 60 kilometres

~ Now what about the displacement!?

Explanation: As he go and came back too, that's why initial position and final position are same as he go to office from home and came back to home again.

→ Displacement = 0 m

Explore more:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}$