Question

A ship is asked to drop a package to the neighboring harbor in a given period of time. The ship travels 20km north first then takes a right and travels 50km and reaches the desired harbor before the allotted time. Calculate the distance and displacement travelled by the ship​

Answer 1

Given data:

The ship travels 20km north first then takes a right and travels 50km and reaches the desired harbor before the allotted time.

To find:

• The distance traveled by the ship.
• The displacement made by the ship.

Solution:

Distance is nothing but the total units traveled by the object. It is a scalar quantity and is independent of the direction of motion.

Distance = The length of the complete path traveled by the object.

Displacement is a vector quantity and gives the shortest distance between the initial and final position of the motion of the object. It is dependent upon the direction of motion.

Displacement = Shortest path of change in position.

To get a clear view between the two as per the given question, check the ATTACHMENT given below.

The ship travels 20km north first then takes a right and travels 50km and reaches the desired harbor before the allotted time.

Distance = 20 km + 50 km = 70 km

Displacement = Shortest path

                        = $\sqrt{20^{2} + 50^{2} }$                 [ Pythagoras rule]

                        = √(400+2500)

                        = √2900

                       = 10√29 km

Therefore, Distance travelled = 70 km, Displacement = 10√29 km

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