Question

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

Answer 1

√29

Step-by-step explanation:

On Khan Academy

Answer 2

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}`

`=√(25+4+0) m` [cos90= 0]

`=√(29) m`

`= 5.38516 m` [ans]