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**Table of Content**

Use this displacement calculator to find the change in position of an object. Our calculator lets you make displacement calculations under various conditions of motion using different formulas:

- If the displacement is based on the average velocity and it remains constant for time (t):

$$S = (av)t$$

- If the displacement involves the object’s initial velocity (u) and the acceleration experienced throughout the time (t):

$$S = ut + \dfrac{1}{2} at^2$$

- If the displacement assumes a constant velocity based on initial and final speeds along with the time:

$$S = \dfrac {1}{2} (v + u) t$$

**“The displacement is the shortest distance, an object travels in a straight line from its starting point to its ending point”**

It represents the change in the position of the object, considering both magnitude and direction.

**Quantity =**Vector quantity**SI Unit =**meter

Here’s an example which perfectly illustrates the concept:

For instance, if a car moves 20 meters east, then 10 meters west, its total distance traveled is 30 meters (20 meters + 10 meters). However, its displacement would be 10 meters east because its final position is 10 meters east from its starting point.

There are several formulas for calculating displacement considering both the distance and direction of an object. There are examples below that helps to indicate these formula throughout the calculations!

If the car moves with an average velocity of \(\ 40ms^{-1}\) for \(\ 120s\), how much does the car cover?

\(\ S = v \times t\)

\(\ S = 40 ms^{-1} \times 120s\)

\(\ S = 4800m\)

A body starts moving from the rest and reaches a velocity of \(\ 80ms^{-1}\) after \(\ 30s\). Find the displacement of the body assuming constant average velocity during this time

- Initial velocity ‘u’ = \(\ 0ms^{-1}\)
- Final velocity ‘v’ = \(\ 80ms^{-1}\)
- Time ‘t’ = \(\ 30s\)

\(\ S = \frac {1}{2}(v + u)t\)

\(\ S = \frac {1}{2}(80 + 0)30\)

\(\ S = \frac{1}{2}(2400)\)

\(\ S = 1200m\)

The car has an initial velocity of \(\ 20ms^{-1}\) and experiences an acceleration \(\ 2ms^{-2}\) for \(\ 15s\). Find the displacement of the car.

- Initial velocity ‘u’ = \(\ 20ms^{-1}\)
- Final velocity ‘v’ = \(\ 2ms^{-2}\)
- Time ‘t’ = 15s

\(\ S = (20)(15) + \frac {1}{2}(2)(15)^2\)

\(\ S = 300 + \frac{1}{2}(2)(225)\)

\(\ S = 300 + \frac{1}{2}(450)\)

\(\ S = 300 + 225\)

\(\ S = 525m\)

Feature | Distance | Displacement |
---|---|---|

Definition | Total length of the path traveled | Straight line path between starting and ending points |

Direction | No | Yes |

Quantity | Scalar (magnitude only) | Vector (magnitude and direction) |

Possible values | Always positive or zero | Positive, negative, or zero |

Path dependence | Depends on the actual path taken | Independent of the path taken |

Measurement | Measured along any path | Measured in a straight line |

Example | A car driving a curvy route for 10 km | A person walking 4 steps east and then 4 steps west, ending at the starting point (displacement of 0 km) |

These are some key applications of the calculator:

**Physics problems:**Determine the traveled distance in various situations of linear motion (1D)**Navigation:**Find the direct route and compass heading between two points on a map (accounting for Earth’s curvature for long distances).**Engineering:**Provide support in calculations related to the motion of an object in various engineering applications.**Everyday Use:**Calculate the direct distance covered, valuable for activities like estimating walking distance or travel paths.

To calculate the displacement without time, measure the final distance and then subtract the starting distance.

**d = |x2 – x1|**

No, the displacement of an object can be either equal to or even less than the distance traveled by the object.