# Distance And Displacement Worksheet

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## Distance And Displacement Worksheet Answers | akademiexcel.com

Distance = 2 miles + 2 miles = 4 miles Displacement = 0 miles because she began and stopped in the similar place. 9. Sandy ran Three blocks north, and then 2 blocks west. Express your solution in blocks. Distance = Three blocks + 2 blocks = Five blocks Because Sandy turns at a proper attitude, we will be able to find her displacement with the Pythagorean Theorem. A2 + B2 = C2Remember: Displacement is the straight-line distance between 2 issues. To give a displacement we will have to give each the dimensions and the route. To find the size of the displacement, count the number of spaces from the initial to the general position. The following displays a displacement of -Five m -7 -5 -4 -3 -1-6 -2 Zero 2135467 Meters fx i fx iPosition, Distance and Displacement worksheet ..\..\bodily science worksheets\introduction to put distance, and displacement lesson.pdf. Distance vs. Displacement Bull's Eye Comparison. CCHS Speedway Lab. Velocity practice problems ..\..\bodily science worksheets\Velocity Problems wksht 1.docThis is a worksheet with solution keys integrated which can assist your newcomers expand their working out of Distance & Displacement. These distinctive workout routines have been designed to assist students:apply their working out of the adaptation between distance and displacement. measure distance and di ### Distance vs Displacement - YouTube

The displacement and distance traveled wouldn't have to be the similar. The runner travels 50 m within the authentic direction (north) plus 30 m in the opposite direction (south), so the entire distance she ran is eighty m. Suppose a runner jogs to the 50-m mark and then turns round and runs again toTitle: Distance and Displacement Worksheet Author: hrsbtech Last modified by: WSFCS Workstation Created Date: 9/4/2013 8:28:00 PM Other titles: Distance and Displacement WorksheetAbout This Quiz & Worksheet. Use this quiz and worksheet to evaluate your knowledge of distance and displacement, and how they vary from each other.Distance and Displacement work sheet. See more. Report this resourceto let us know if it violates our terms and conditions. Our customer service staff will review your file and will be in touch.The most comprehensive library of loose printable worksheets & virtual video games for children. Get thousands of teacher-crafted actions that sync up with the school yr.

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## Distance and Displacement-

Before you undergo this text, just be sure you have gone through the previous article on Distance and Displacement.

We have discussed-

Distance is the total period of the actual trail travelled by a particle during its movement. Displacement is a vector drawn from the initial place to the overall position of the particle.

In this newsletter, we will be able to discuss issues 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-

2πR, 0 2R, πR πR/2, 2R πR, R

## Solution-

Consider particle moves from level A to point B protecting 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 coated by way of the particle

= Total duration of the actual trail

= Circumference of the circle / 2

= 2πR / 2

= πR gadgets

Thus, Option (B) is proper.

## Problem-02:

A particle strikes in 1 / 4 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 strikes from point A to indicate B covering half of the circle as shown-

Distance-

Distance covered via the particle

= Total duration of the particular path

= Length of arc AB

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

= πR/2 devices

Displacement-

Magnitude of displacement of the particle

= Shortest distance between point A and level B

= Length of line AB

= √(OA2 + OB2)

= √(R2 + R2)

= √2R units

Thus, Option (A) is correct.

## Problem-03:

A mosquito flies from some extent A(-1, 2, 3) m to some extent 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-

||

= √(32 + 32 + 52)

= √(9 + 9 + 25)

= √43 m

Thus, magnitude of displacement of the mosquito = √Forty three m.

## Problem-04:

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

## Solution-

Consider the beginning of Cartesian coordinate system to be provide at point A.

Then-

Coordinates of level 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-

||

= √(52 + 52 + 52)

= √(3 x 52)

= 5√3 m

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

## Problem-05:

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

πR/3, R πR/6, √2R πR/3, √2R πR/3, √2R

## Solution-

Consider particle strikes from point A to indicate B overlaying an angle of 60º from its initial position as shown-

Distance-

Distance lined via the particle

= Total period of the particular path

= Length of arc AB

= π/Three x R        Using the relation θ = arc / radius

= πR/Three gadgets

Displacement-

In Δ AOB,

OA = OB = R ∠AOB = 60°

Now,

We know, angles reverse to equal facets 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 of the angles in ΔAOB are 60°, so ΔAOB is an equilateral triangle.

So, all of the sides length must be same.

Thus, period of line AB = R gadgets.

Now,

Magnitude of displacement of the particle

= Shortest distance between level A and level B

= Length of line AB

= R units

Thus, Option (A) is proper.

## Problem-06:

A participant completes a circular path of radius R in forty seconds. Its distance and displacement on the finish of 2 minutes 20 seconds will be-

7πR, 2R 2R, 2R 2πR, 2R 7πR, R

## Solution-

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

So, choice of revolutions made via the participant 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.Five x 2πR

= 7πR gadgets

Displacement-

Magnitude of displacement after 3.Five revolutions

= Diameter of the circle

= 2R gadgets

Thus, Option (A) is right kind.

## Problem-07:

Given that P is a point on a wheel rolling on horizontal highway. The radius of the wheel is R. Initially, the purpose P is involved with floor. The wheel rolls via part of the revolution. What is the displacement of level P?

## Solution-

Given-

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

After rolling through half revolution, level P is at the top most level 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 issues on distance and displacement,

Watch this Video Lecture

Next Article- Speed and Velocity

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

Get more notes and other learn about subject matter of Class 12 Physics.

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