Question

A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?

Answer 1

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

           Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

                            $R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}$

                            $=\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}$

                            $=\sqrt{25+4+0} m$  [cos90= 0]

                            $=\sqrt{29} m$

                            $= 5.38516 m$   [ans]