## Distance and Displacement-

Before you go through this article, make sure that you have gone through the previous article on Distance and Displacement.

We have discussed-

• Distance is the total length of the actual path travelled by a particle during its motion.
• Displacement is a vector drawn from the initial position to the final position of the particle.

## Problem-01:

A particle covers half of the circle of radius R. Then, the displacement and distance of the particle are respectively-

1. 2πR, 0
2. 2R, πR
3. πR/2, 2R
4. πR, R

## Solution-

Consider particle moves from point A to point B covering half of the circle as shown-

### Displacement-

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Diameter of the circle

= 2R units

### Distance-

Distance covered by the particle

= Total length of the actual path

= Circumference of the circle / 2

= 2πR / 2

= πR units

Thus, Option (B) is correct.

## Problem-02:

A particle moves in a quarter circle of radius R. What are the values of distance and displacement respectively?

1. πR/2, √2R
2. πR, R
3. 2√2R, πR/2
4. πR/2, 0

## Solution-

Consider particle moves from point A to point B covering half of the circle as shown-

### Distance-

Distance covered by the particle

= Total length of the actual path

= Length of arc AB

= π/2 x R       { Using the relation θ = arc / radius }

= πR/2 units

### Displacement-

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Length of line AB

= √(OA2 + OB2)

= √(R2 + R2)

= √2R units

Thus, Option (A) is correct.

## Problem-03:

A mosquito flies from a point A(-1, 2, 3) m to a point B(2, -1, -2) m. Find the displacement of the mosquito.

## Solution-

We have-

• Initial position vector, $\overrightarrow{r1}&space;=&space;-\hat{i}&space;+&space;2\hat{j}&space;+&space;3\hat{k}$
• Final position vector, $\overrightarrow{r2}&space;=&space;2\hat{i}&space;-&space;\hat{j}&space;-&space;2\hat{k}$

Displacement vector of the mosquito $\overrightarrow{r}$ is given by-

$\overrightarrow{r}&space;=&space;\overrightarrow{r2}&space;-&space;\overrightarrow{r1}$

So, we have-

$\overrightarrow{r}&space;=&space;(2\hat{i}&space;-&space;\hat{j}&space;-&space;2\hat{k})-(-\hat{i}&space;+&space;2\hat{j}&space;+&space;3\hat{k})$

$\overrightarrow{r}&space;=&space;3\hat{i}&space;-&space;3\hat{j}&space;-&space;5\hat{k}$

Thus, displacement vector of the mosquito is-

${\color{DarkRed}\mathbf{\overrightarrow{r}&space;=&space;3\hat{i}&space;-3\hat{j}&space;-&space;5\hat{k}}}$

The magnitude of displacement is given by-

|$\overrightarrow{r}$|

= √(32 + 32 + 52)

= √(9 + 9 + 25)

= √43 m

Thus, magnitude of displacement of the mosquito = √43 m.

## Problem-04:

A mosquito flies from corner A to the corner B of a cube of side length 5 m moving along its sides as shown. Find the displacement of the mosquito.

## Solution-

Consider the origin of Cartesian coordinate system to be present at point A.

Then-

• Coordinates of point A = (0, 0, 0)
• Coordinates of point B = (5, 5, 5)

Displacement vector of the mosquito is given by-

$\overrightarrow{r}&space;=&space;(5-0)\hat{i}&space;+&space;(5-0)\hat{j}&space;+&space;(5-0)\hat{k}$

$\overrightarrow{r}&space;=&space;5\hat{i}&space;+&space;5\hat{j}&space;+&space;5\hat{k}$

Thus, displacement vector of the mosquito is-

${\color{DarkRed}&space;\mathbf{\overrightarrow{r}&space;=&space;5\hat{i}&space;+&space;5\hat{j}&space;+&space;5\hat{k}}}$

The magnitude of displacement is given by-

|$\overrightarrow{r}$|

= √(52 + 52 + 52)

= √(3 x 52)

= 5√3 m

Thus, magnitude of displacement of the mosquito = 5√3 m.

## Problem-05:

A particle is moving in a circular path of radius R. The distance and displacement of the particle when it describes an angle of 60° from its initial position respectively are-

1. πR/3, R
2. πR/6, √2R
3. πR/3, √2R
4. πR/3, √2R

## Solution-

Consider particle moves from point A to point B covering an angle of 60º from its initial position as shown-

### Distance-

Distance covered by the particle

= Total length of the actual path

= Length of arc AB

= π/3 x R       { Using the relation θ = arc / radius }

= πR/3 units

### Displacement-

In Δ AOB,

• OA = OB = R
• ∠AOB = 60°

Now,

• We know, angles opposite to equal sides are equal. So, ∠OAB = ∠OBA.
• Also, we know sum of three angles in a triangle is 180°.
• So, we conclude ∠OAB = ∠OBA = 60°.

Since all the angles in ΔAOB are 60°, so ΔAOB is an equilateral triangle.

So, all the sides length must be same.

Thus, length of line AB = R units.

Now,

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Length of line AB

= R units

Thus, Option (A) is correct.

## Problem-06:

A player completes a circular path of radius R in 40 seconds. Its distance and displacement at the end of 2 minutes 20 seconds will be-

1. 7πR, 2R
2. 2R, 2R
3. 2πR, 2R
4. 7πR, R

## Solution-

We have number of revolutions made by the player in 40 seconds = 1.

So, number of revolutions made by the player in 2 minutes 20 seconds (140 seconds)

= $\frac{1}{40}$ x 140

= 3.5 revolutions

### Distance-

Distance covered in 3.5 revolutions

= 3.5 x Distance covered in 1 revolution

= 3.5 x Circumference of the circle

= 3.5 x 2πR

= 7πR units

### Displacement-

Magnitude of displacement after 3.5 revolutions

= Diameter of the circle

= 2R units

Thus, Option (A) is correct.

## Problem-07:

Given that P is a point on a wheel rolling on horizontal road. The radius of the wheel is R. Initially, the point P is in contact with ground. The wheel rolls through half of the revolution. What is the displacement of point P?

## Solution-

Given-

• Initially, the point P is in contact with ground.
• The wheel rolls through half of the revolution.

After rolling through half revolution, point P is at the top most point of the wheel.

So, we have-

Here, d represents the displacement of point P.

Using Pythagoras theorem, we have-

d2 = (2R)2 + (πR)2

d2 = 4R2 + π2R2

d2 = (4+π2)R2

∴ d = R√(4+π2)

Thus, magnitude of displacement of point P = R√(4+π2) m.

To practice more problems on distance and displacement,

Watch this Video Lecture

Next Article- Speed and Velocity

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## Motion in One Dimension-

Before you go through this article, make sure that you have gone through the previous article on Motion in One Dimension.

We have discussed-

• In one-dimensional motion, particle moves along a straight line.
• It is also known as rectilinear motion.

Various parameters related to motion are-

## Distance-

 Distance is the total length of the actual path travelled by a particle during its motion.

### Example-

Consider-

• A particle moves in a circular path of radius R.
• It starts its journey from point A and ends at the same point.

Distance covered by the particle during its motion

= Circumference of the circle

= 2πR units

### Characteristics-

The characteristics of distance are-

• It depends upon the path followed by the particle.
• It is a scalar quantity.
• Distance is always positive. It can never be negative.
• Distance can not decrease with time.
• Distance is never zero for a moving particle.
• The SI unit of distance is meter (m).
• The dimensional formula of distance is [M0L1T0].

## Displacement-

 Displacement is a vector drawn from the initial position to the final position of the particle. Its magnitude is equal to the shortest distance between the initial and final position.

Mathematically, it is equal to the change in position vector of the particle.

Consider-

• A particle is initially at point A having position vector $\overrightarrow{r1}$.
• The particle moves to point B having position vector $\overrightarrow{r2}$.

### Example-

Consider-

• A particle moves along a straight line from point O to point A.
• It then travels back from point A to point B as shown-

Magnitude of displacement of the particle

= Shortest distance between its initial and final position

= 70 m – 0 m

= 70 m

Direction of displacement is towards +ve X-axis.

It is important to note that here the distance covered by the particle = 100 m + 30 m = 130 m.

### Characteristics-

The characteristics of displacement are-

• It is independent of the path followed by the particle.
• It is a vector quantity.
• Displacement can be positive, negative or zero.
• Displacement can decrease with time.
• Displacement may be zero for a moving particle.
• The SI unit of displacement is meter (m).
• The dimensional formula of displacement is [M0L1T0].

## Difference Between Distance And Displacement-

Some important differences between distance and displacement are-

 Distance Displacement It is the total length of the actual path travelled by a particle during its motion. It is the shortest distance between the initial and final positions. It depends upon the path followed by the particle. It is independent of the path followed by the particle. It is a scalar quantity. It is a vector quantity. Distance is either positive or zero. It can never be negative. Displacement can be positive, negative or zero. Distance can never decrease with time. Displacement may decrease with time. Distance between a given set of initial and final position can have infinite values. Displacement between a given set of initial and final position is unique. Distance is never zero for a moving particle. Displacement may be zero for a moving particle.

## Important Points-

It is important to note the following points-

### Point-01:

• Magnitude of displacement can never be greater than distance.
• Magnitude of displacement is equal to the distance if the particle moves along a straight line without changing the direction.

### Point-02:

When a particle returns to its initial position,

• Its displacement is always zero.
• Its distance covered is not zero.

### Point-03:

According to convention,

• If the particle moves upwards or towards right, its displacement is taken positive.
• If the particle moves downwards or towards left, its displacement is taken negative.

To gain better understanding about Distance and Displacement,

Watch this Video Lecture

Next Article- Problems On Distance & Displacement

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