Distance and Displacement Problems | Worksheet

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-

  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} = -\hat{i} + 2\hat{j} + 3\hat{k}
  • Final position vector, \overrightarrow{r2} = 2\hat{i} - \hat{j} - 2\hat{k}

 

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

\overrightarrow{r} = \overrightarrow{r2} - \overrightarrow{r1}

 

So, we have-

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

\overrightarrow{r} = 3\hat{i} - 3\hat{j} - 5\hat{k}

 

Thus, displacement vector of the mosquito is-

{\color{DarkRed}\mathbf{\overrightarrow{r} = 3\hat{i} -3\hat{j} - 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} = (5-0)\hat{i} + (5-0)\hat{j} + (5-0)\hat{k}

\overrightarrow{r} = 5\hat{i} + 5\hat{j} + 5\hat{k}

 

Thus, displacement vector of the mosquito is-

{\color{DarkRed} \mathbf{\overrightarrow{r} = 5\hat{i} + 5\hat{j} + 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

 

Get more notes and other study material of Class 11 Physics.

Get more notes and other study material of Class 12 Physics.

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Distance and Displacement Problems | Worksheet
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Distance and Displacement Problems | Worksheet
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Distance and Displacement Problems- We discuss questions with solutions for practice on distance and displacement. Distance and Displacement Worksheet With Answers.
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