Question

A body travels a distance of 3 km towards east then 4 km towards north and finally 9 km towards east (a) what is the total distance travelled ?
(b) what is the resultant displacement

Answer 1

Solution :

⏭ Given:

✏ A body travels a distance of 3km towards east then 4km towards north and finally 9km towards east.

⏭ To Find:

• Total Distance
• Total Displacement

⏭ Concept:

✏ Distance is defind as total lengh of path which is covered by body.

✏ Displacement is defind as the shortest distance between two points.

⏭ Calculation:

• Total Distance

✏ Distance = 3 + 4 + 9

✏ Distance = 16 km

• Total Displacement

$\implies \sf \: s = \sqrt{ {12}^{2} + {4}^{2} } \\ \\ \implies \sf \: s = \sqrt{144 + 16} = \sqrt{160}$

✏ Displacement ≈ 12.64 km

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

A body travels 3 km towards east then 4 km towards north and finally 9 km towards east

See the attachment !

We know that , the length of path traveled by body is called distance

Thus ,

Distance = 3 + 4 + 9

Distance = 16 km

Hence , the distance traveled by the body is 16 km

And the shortest distance b/w final and initial position is called displacement

Thus ,

By Pythagoras theorem ,

$\sf \mapsto AD = \sqrt{ {(12)}^{2} + {(4)}^{2} } \\ \\ \sf \mapsto AD = \sqrt{144 + 16} \\ \\ \sf \mapsto AD = \sqrt{160} \\ \\ \sf \mapsto AD = 12.6 \: \: km$

Hence , the displacement is 12.6 km

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