# Numericals on distance and displacement class 9

In this post, let’s solve a set of Numericals on distance and displacement for class 9.

**Concepts Used | Formulas Used**

Distance

The distance is the length of the actual path traversed by an object irrespective of the direction in which the object travels.

It is a scalar quantity and it can never be zero or negative during the motion of an object.

Its SI unit is meter.

Displacement

The shortest distance between the initial and final positions of any object during motion, along with direction, is called displacement.

The displacement of an object in a given time can be positive, zero, or negative.

It is a vector quantity. Its unit is meter.

Displacement = (Final position – Initial position)

ΔX = (X2-X1)

**Numericals on distance and displacement for class 9**

Question 1] A body is moving along a circular path of radius R. What will be the distance travelled and displacement of the body when it completes half a revolution?

**Solution:**

Circumference of the circle or perimeter of the circle =2πR

Half a revolution means half of the perimeter of the circle i.e. πR

So the distance traveled = πR

For a half-revolution travel, Displacement =diameter of the circle = 2R

Question 2] If on a round trip you travel 6 km and then arrive back home :

(a) What distance have you travelled?

(b) What is your final displacement?

**Solution:**

a) distance travelled = 6 km

b) Here, final position = initial position.

Hence, displacement = 0

Question 3] A body travels a distance of 3 km towards the East, then 4 km towards the North, and finally 9 km towards the East.

(i) What is the total distance travelled?

(ii) What is the resultant displacement?

**Solution:**

i) Total distance travelled = (3 + 4 + 9) km = 16 km

ii) Let’s draw a supportive diagram.

In this diagram, O is the starting point. OA = 3 km, AB = 4 km, and BC = 9 km.

Hence, AX = BC = 9 km

OX = OA + AX = (3 + 9) km = 12 km

and, CX = AB = 4 km

In the triangle OCX, we have to find out the value of OC as its the representation of the displacement for this scenario. OC = [OX^{2} + CX^{2}]^{1/2} = [12^{2} + 4^{2}]^{1/2} = 12.64

Displacement = 12.64 m

Question 4] A man moves 3 km towards the East then moves 5 km towards the South and finally 9 km towards the North. How far is he from the starting point?

Solution:

Let’s draw a supportive diagram. In this diagram, O is the starting point. OA = 3 km, AB = 5 km, BC = 9 km

So, AC = (9-5)km = 4 km

In the OAC triangle, OA = 3 km, AC =4 km

So, OC = displacement of the man = [OA^{2}+ AC^{2}]^{1/2}= [3^{2}+ 4^{2}]^{1/2}= 5 km

So, the man is 5 km away from the starting point.

**Question 5] An object is moving along a circular path of radius R. What will be the distance travelled and displacement of the body when it completes a revolution?**or

**Calculate the distance and displacement of a body that moves a complete circle**

**of radius**R.

**Solution**:

Circumference of the circle or perimeter of the circle =2πR

One revolution means traveling the perimeter of the circle i.e. 2πR

So the distance traveled = 2πR

For a full revolution trip, the final position equals the initial position. Hence, Displacement =0

**Question 6] A farmer moves along the boundary of a square field of side 15 m in 30 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 15 seconds?** **Also, find out the distance travelled by the farmer.**

**Solution:**

Here is a supportive diagram where ABCD is a square field of side 15 m.

Time for one round of the square field = 30 s

Total time taken = 2 min 15 s = (2 x 60 + 15) s = 135 s

Number of rounds completed = 135/30 = 4.5

This means, if the farmer starts from A after 4.5 rounds, he will be at C.

So, displacement of the farmer = AC = [AB^{2}+ BC^{2}]^{1/2}= [15^{2}+ 15^{2}]^{1/2}= 21.21 m

Each round of traversal means travelling the total length of the 4 sides of the square field = 4 x 15 m = 60 m.

So distance travelled by the farmer after 4.5 rounds = 60 x 4.5 m = 270 m