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A body goes to $10\,km$ north and $20\,km$ east. The displacement from initial point is
A. $22.36\,km$
B. $2\,km$
C. $5\,km$
D. $20\,km$

Answer
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Hint: The displacement of an object in a given interval of time is defined as the shortest distance between the two positions of the object in a particular direction during that time and is given by the vector drawn from the initial position to its final position.

Complete step by step answer:
Let the body go along OA = 10 km towards north and OB = 20 km towards east. Then the displacement of the body is given by the hypotenuse AB of the triangle OAB.
Displacement = $\sqrt {{{10}^2} + {{20}^2}} $ = $\sqrt {100 + 400} = \sqrt {500} $= 22.36 km
Hence the correct option is (A).

Note:The length of the actual path traversed by an object during motion in a given interval of time is called the distance travelled by the object. On the other hand, the displacement of an object in a given interval of time is defined as the shortest distance between the two positions of the object in a particular direction during that time and is given by the vector drawn from the initial position to its final position. Displacement is a vector quantity and possesses both direction and magnitude. When an object goes along the path \[\overrightarrow {AB} \] , the arrowhead at B shows that the object is displaced from A to B. In case the object is displaced from B to A, it is represented by \[\overrightarrow {BA} \]. It means that the magnitude of \[\overrightarrow {AB} \] is the same as that of \[\overrightarrow {BA} \] but opposite in direction.
Last updated date: 04th Jun 2023
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