Question

a boy travels 5km towards north then he turns right and travels another 5km before coming to rest. Calculate total distance and total displacement.

Answer 1

GiveN :-

A boy travels 5 km towards north then he turns right and travels another 5 km before coming to rest.

TO FinD :-

The total distance and total displacement.

SolutioN :-

We know that, Distance is defined as the amount of space between two things or the total length of the path travelled between two objects.

♣ The total distance travelled = 5 km + 5 km = 10 km

$\rule{200}{2}$

For displacement (Kindly refer to the attachment for the diagram) :-

Using Pythagoras Theorem,

Displacement² = 5² + 5²

⇒Displacement = √(25 + 25)

⇒Displacement = √50

♣ ⇒Displacement = 5√2 km

Answer 2

Given:

• A boy travels 5km towards north
• Then he turns right and travels another 5km before coming to rest

To Find:

• Distance travelled by boy
• Displacement of boy

Diagram:

See in the attachment

Solution:

We know that,

• Distance is the length of actual path covered by body
• Displacement is the distance between starting and final position
• According to Pythagoras Theorem, for a right angled triangle

$\pink{\boxed{\rm{(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}}}}$

$\rule{190}{1}$

Now, on applying Pythagoras theorem in ΔABC

$\longrightarrow\rm{(AC)^{2}=(AB)^{2}+(BC)^{2}}$

$\longrightarrow\rm{(AC)^{2}=(5)^{2}+(5)^{2}}$

$\longrightarrow\rm{(AC)^{2}=25+25}$

$\longrightarrow\rm{(AC)^{2}=50}$

$\longrightarrow\rm{AC=\sqrt{50}}$

$\longrightarrow\rm{AC=5\sqrt{2}}$

From the diagram, it is clear that

$\longrightarrow\rm{Distance\:covered\:by\:Boy=AB+BC}$

$\longrightarrow\rm{Distance\:covered\:by\:Boy=5+5}$

$\longrightarrow\rm{Distance\:covered\:by\:Boy=10\:Km}$

Again from diagram, it can be clearly seen that

$\longrightarrow\rm{Displacement\:of\:Boy=AC}$

$\longrightarrow\rm{Displacement\:of\:Boy=5\sqrt{2}\:Km\:(North-East)}$

North-East because direction of displacement is in the direction of north-east with respect to starting position.

Hence, the total distance covered by boy is 10 Km and displacement of boy is 5√2 Km (N-E).