Displacement is shortest distance between initial and final point which prefers straight line path over curved paths.

If a body is moving in two different directions x and y then Resultant displacement is

It gives the short cut paths for the given original paths. Generally it is also given by :

Here,

u is the Initial velocity

If a body is moving in two different directions x and y then Resultant displacement is

u is the Initial velocity

v is the final velocity

a is acceleration

t is the time taken.

While solving the problems if initial and final velocity both are given we use the first formula if final velocity and time taken are given you can use second formula and If you are interested in finding area under the curve then use the third formula.

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Displacement Calculator | Acceleration Formula Calculator |

Area of a Circle Formula Calculator | Area of a Cylinder Formula Calculator |

Solved problems on displacement which helps you to understand the concept better are given below

### Solved Examples

**Question 1: **The path from garden to a school is 5m west and then 4m south. A constructor wants to build a short distance path for it. Help him out?

** Solution: **

Given: Distance to the west x = 5m

Distance to the south y = 4m.

Displacement is given by S = $\sqrt{x^{2} + y^{2}}$

= $\sqrt{5^{2} + 4^{2}}$

= 6.70 m.

The constructor can build a path for displacement length of 6.7 m.

**Question 2: **A girl walks from the corridor to the gate she moves 3m to the north opposite from her house then takes a left turn and walks for 5m. then she takes right turn and moves for 6m and reaches the gate. What is the displacement, magnitude and distance covered by her?

** Solution: **

Total distance traveled d = 3m + 5m + 6m = 14m.

Magnitude of the displacement can be obtained by visualizing the walking :The actual path from A to B as 3m then from B to D as 5m and finally from D to E as 6m.

So the magnitude of resultant displacement is |S| = $\sqrt{AC^{2} + CE^{2}}$

From figure AC = AB + BC = 3m + 6m = 9m

BD = CE = 5m

|S| = $\sqrt{9^{2} + 5^{2}}$ = 10.29 m.

The direction of Resultant displacement is South East.

Given: Distance to the west x = 5m

Distance to the south y = 4m.

Displacement is given by S = $\sqrt{x^{2} + y^{2}}$

= $\sqrt{5^{2} + 4^{2}}$

= 6.70 m.

The constructor can build a path for displacement length of 6.7 m.

Total distance traveled d = 3m + 5m + 6m = 14m.

Magnitude of the displacement can be obtained by visualizing the walking :The actual path from A to B as 3m then from B to D as 5m and finally from D to E as 6m.

So the magnitude of resultant displacement is |S| = $\sqrt{AC^{2} + CE^{2}}$

From figure AC = AB + BC = 3m + 6m = 9m

BD = CE = 5m

|S| = $\sqrt{9^{2} + 5^{2}}$ = 10.29 m.

The direction of Resultant displacement is South East.

More topics in Displacement Formula | |

Angular Displacement Formula | |