Question

a car travel East on level road for 3 km and then turns the north at insertion and travel 4 km before stopping the magnitude of resultant displacement of a car is.​

Answer 1

Anѕwєr :

$\Huge\boxed{\purple{\sf{ 5 \; km }}}$

Explαnαtion :

As per the provided information in the given question, we have :

• A car travels due east on level road for 3 km and then turns the north at intersection and travel 4 km before stopping.

We've been asked to calculate the magnitude of resultant displacement.

$\large {\underline { \sf {Displacement :}}}$

• Displacement is the shortest distance from the body's initial to final position. It is a vector quantity that means it requires both magnitude and direction for its description. Its SI unit is metre.

Suppose the body travels the distance AB that is 3 km due East and then the body travels the distance BC that is 4 km due North.

As the displacement is the shortest distance from initial to final position, so here are the shortest distance from A to C is the straight line path AC.

$\implies\sf{Displacement = AC }$

By using Pythagoras property,

$\implies\sf{AC = \sqrt{(AB)^2 + (BC)^2} }$

$\implies\sf{AC = \sqrt{(3 \; km)^2 + (4\; km)^2} }$

$\implies\sf{AC = \sqrt{9 \; km^2 + 16\; km^2} }$

$\implies\sf{AC = \sqrt{25 \; km^2} }$

$\implies\sf{AC = 5 \; km }$

$\implies\boxed{\purple{\sf{ Displacement =5 \; km }}}$

∴ The magnitude of the resultant displacement of the car is 5 km.

$\rule{200}2$