In this question, we have to find the magnitude of AC Vector

**Options**

- A. 223,7m, tan
^{− 1}( 2 ) , N of E - B. 223,7m, tan
^{− 1 }( 2 ) , E of N - C. 300m, tan
^{− 1}( 2 ) , N of E - D. 100m, tan
^{− 1 }( 2 ) , N of E

**Solution **

Given That the car makes a displacement of 100 m towards east and then 200 m towards north. What is the magnitude and direction of the resultant?

- Basic assumptions,
- Let D denotes the magnitude of resultant displacement AC
- Let θ denotes the angle between the resultant displacement and the displacement of 100 m towards east.

- Hence from above data we get following relations,
- D = √(100^2 + 200^2)
**or**D = 100*√(1 + 4) = 100*√5 = 100*2.236 = 223.6 (m)

- D = √(100^2 + 200^2)

- tanθ = 200/100
**or**tanθ = 2**or**θ = tan^**–**1 (2) = 63.43°**[Ans]**

### What is displacement?

Displacement is the shortest distance from point A to point B