Question

An object moves 6 km due west and then 8 km due
morth. Find the distance covered by the object and its displacement.​

Answer 1

Solution :

★ Displacement -

Displacement is the shortest distance from the origin to the final point. According to the question, (refer the attachment) AC forms the shortest distance.

By Pythagoras Theorem,

In ∆ ABC,

>> Hypotenuse = CA = y

>> Height = BC = 8 km

>> Base = BA = 6 km

$\bigstar \: {\boxed{\sf{(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}}}}$

⇒ (y)² = (6)² + (8)²

⇒ y² = 36 + 64

⇒ y² = 100

⇒ y = √100

⇒ y = 10

CA = 10 km

Shortest distance = 10 km

Therefore, the displacement is 10 km.

$\rule{300}{1.5}$

★ Distance -

⇒ 8km + 6km

⇒ 14 km

Therefore, the distance covered is 14 km.

Answer 2

Answer:-

Given:-

• First, it moves 6 km Westwards.
• Second, it moves 8 km Northwards.

To Find:-

• Distance covered.
• Displacement.

_________________...

Diagram:-

'N'

• ╲

| ╲

| ╲

8 | ╲

km | ╲

| ╲

|_ 90° ╲

|_ | _________________ •

'W' 6 kms 'O'

• Let O be the initial position of the object.
• W be the end after covering 6 km westwards.
• N be the end after covering 8 km northwards.

Since, distance = measurement of actual path covered,

∴ Distance = OW + WN

= 6 km + 8 km

= 14 km ...(Ans. 1)

NOW, displacement:-

Displacement = Shortest distance (straight line) between the initial and the final position of a particle.

∴ Displacement = ON

By applying Pythagoras Theorem,

[(hypo.)² = (base)² + (altitude)²]

∴ (ON)² = (6)² + (8)²

➵ (ON)² = 36 + 64

➵ ON = √100

➵ ON = 10 kms ...(Ans. 2)

∴ Distance covered in the journey was 14 km and the displacement was 10 km.