**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.

**PRACTICE PROBLEMS BASED ON DISTANCE & DISPLACEMENT-**

**Problem-01:**

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

- 2πR, 0
- 2R, πR
- πR/2, 2R
- 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?

- πR/2, √2R
- πR, R
- 2√2R, πR/2
- 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

= √(OA^{2} + OB^{2})

= √(R^{2} + R^{2})

= √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,
- Final position vector,

Displacement vector of the mosquito is given by-

So, we have-

Thus, displacement vector of the mosquito is-

The magnitude of displacement is given by-

||

= √(3^{2} + 3^{2} + 5^{2})

= √(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-

Thus, displacement vector of the mosquito is-

The magnitude of displacement is given by-

||

= √(5^{2} + 5^{2} + 5^{2})

= √(3 x 5^{2})

= 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-

- πR/3, R
- πR/6, √2R
- πR/3, √2R
- 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-

- 7πR, 2R
- 2R, 2R
- 2πR, 2R
- 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)

= 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-

d^{2} = (2R)^{2} + (πR)^{2}

d^{2} = 4R^{2} + π^{2}R^{2}

d^{2} = (4+π^{2})R^{2}

∴ d = R√(4+π^{2})

Thus, magnitude of displacement of point P = **R√(4+π ^{2}) m**.

To practice more problems on distance and displacement,

**Next Article-** **Speed and Velocity**

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