Question

A boy moves along the path ABCD. What is the total distance Covered by the boy? What is his net displacement?

Answer 1

distance= ab+bc+cd

=50+40+20=110

displacement=4.721

Answer 2

Answer:

The distance covered is 110 meters.

The net displacement is 50 meters.

Explanation:

Distance:

Distance is the length of the path traveled by an object to reach from one point to another point. Distance is a scalar quantity.

AB = 50 m

BC = 40 m

CD = 20 m

The total distance = AB + BC + CD

                                     = 110 m

Displacement:

Displacement is the length of the shortest distance between two points. Displacement is a vector quantity.

Let AE = AB - CD

             = 50m - 20m

             = 30 m

Δ AED is a right-angle triangle,

AD is the shortest distance between two points, A and D.

$AD^{2}$ = $AE^{2}$ + $ED^{2}$

$AD^{}$ = $\sqrt{AE^{2} + ED^{2}  }$

$AD^{}$ = $\sqrt{30^{2} + 40^{2}  }$

$AD^{}$ = $\sqrt{250}$

AD = 50 m

The net displacement is 50 m