## 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.

In this article, we will discuss problems on distance and displacement.

## 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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