A person moves 30\ m North and then 20\ m towards East and finally 30\sqrt{2} m in South-West direction. The displacement of the person from the origin will be

Correct Answer: Option CHint: By drawing a rough figure we can find the direction of displacement of the person, magnitude can be found by using the Pythagoras theorem.Explanation of Correct Option:Let AB be the displacement of the person towards the North, BC be the displacement towards the south and CE be the displacement in the South-West direction. We have to find the displacement of the person from the origin, EA Consider the right triangle ECD , by Pythagoras theorem ED^{2}+CD^{2}=EC^{2}\rightarrow(i) From fig, CD=AB=30m EC=30\sqrt{2}m ED=EA+AD AD=BC=20m \therefore ED= EA+20m (i)\rightarrow(EA+20)^{2}+(30)^{2}=(30\sqrt{2})^{2} (EA+20)^{2}+900=1800 EA+20=30 EA=10m The displacement of the person from the origin will be 10m along West.

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>>Class 11
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>>Distance and Displacement
>> $Uniform and Non-Uniform Motion$

Uniform and Non-Uniform Motion

Q: An athlete completes one round of a circular track of diameter 200 m in 40 s. If he continues to run, what will be the distance covered and the displacement respectively, at the end of 2 minutes 20 s?

The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = 2\pi r\times 3.5=2\times 3.14\times 100\times 3.5=2200\ m .For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.Hence, the distance covered is 2200 m and the displacement is 200 m.

Q: A farmer moves along the boundary of a square field of side 10 m in 40 sec What will be magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Given side of square =10 m , thus perimeter P=40 m Time taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = 1\times 140=140 m Now the total number of rotation the farmer makes to cover a distance of 140 meters = \dfrac{total\quad distance }{perimeter} =3.5 At this point, the farmer is at a point say B from the origin OThus the displacement s=\sqrt{{10}^{2}+{10}^{2}} from Pythagoras theorem.s= 10\sqrt{2}=14.14 m

Q: A particle moves along a straight line OX. At a time t(in seconds) the distance x(in meters)of the particle from 0 is given by:- x=40+12t-t^3 . How long would the particles travel before coming to rest?

At t=0 , particle is at , let's say x distance ,from O ; then putting t=0 in the given displacement-time equation we get ; x =40 +12(0) -(0)^3 = 40 m Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement ; let's say the time be t. then after differentiating the given displacement - time equation wrt. time we get velocity - time equation v= 12-3t^2 at time t =t ( the time when the particle comes to rest ): v= 0; => 12-3t^2 = 0; => t = 2 s Then , at t =2s we are at , let's say x' distance from O ; put this value of t (=2) in given displacement-time equation , we get ; x'= 40 +12(2) -(2)^3; = 56m Further ; We have seen that the particle started his journey when it is at 40m from the point O. And came to rest at 56m from the point O . then the particle traveled a distance of : 56-40=16m. Hence,option (A) is correct answer.

Q: The numerical ratio of displacement to distance for a moving object is

We know, magnitude of displacement is the staright line distance between the initial point and the final point of the path of the moving body.Dispalcement could be negative, positive or zero. but distance is always positive or zero when there is no motion.So, the ratio of displacement to distance for a moving object is equal to less than 1.

Q: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

v=60km/h=\cfrac{60\times 1000}{3600}m/s t=9s Length of the pole =vt\\=\cfrac{60\times 1000}{3600}\times 9\\=150m

Q: Distinguish between Distance and displacement.

DistanceDisplacement 1. It is the actual length of path followed by the body while moving between two positions. 1. It is the linear distance between the initial and final positions of a body. 2. It has only magnitude - it is a scalar quantity. 2. It has both magnitude and direction - it is a vector quantity. 3. Its SI unit is m. 3. Its SI unit is m.

Q: If the distance between earth and the sun were half of its present value , then how many number of days will be there in at year?

According to Kepler’s law T² ∝ r³ (T₁ / T₂)² = (r₁ / r₂)³ (365 / T₂)² = (2r₁ / r₁)³ (365 / T₂)² = 8 365 / T₂) = 2√2 T₂ = 365 / (2√2) T₂ = 129 days There will be 129 days in an year.

Q: Two cars start from opposite places on the main road, 150 km apart. The first car runs from 25 km and takes a right turn and then runs 15 km. It then turns left and runs for another 25 km and then takes the direction to reach the main road. In the meantime, due to a minor break down the other car has run only 35 km along the main road. What would be the distance between the two cars at this instant?

Answer is A.Let us consider the below diagrammatic representation of the case given in the question.The required distance is = DF.That is, DF = 150 - (25 + 25 +35) = 150 - 85 = 65 km.Hence, the distance between the two cars at that point is 65 km.

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Two cars start from opposite places on the main road, 150 km apart. The first car runs from 25 km and takes a right turn and then runs 15 km. It then turns left and runs for another 25 km and then takes the direction to reach the main road. In the meantime, due to a minor break down the other car has run only 35 km along the main road. What would be the distance between the two cars at this instant?

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Uniform and Non-Uniform Motion

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REVISE WITH CONCEPTS

Introduction to Motion

Understanding Uniform and Non-Uniform Motion

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Uniform And Non-Uniform Velocity

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Basics of Motion

Distance and Displacement

Understanding Uniform Motion

Understanding Non-Uniform Motion

Basic Knowledge of Uniform and Non-Uniform Motion and Average Speed

Important Questions

An athlete completes one round of a circular track of diameter 200 m in 40 s. If he continues to run, what will be the distance covered and the displacement respectively, at the end of 2 minutes 20 s?

A farmer moves along the boundary of a square field of side 10 m in 40 sec What will be magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

A particle moves along a straight line OX. At a time t(in seconds) the distance x(in meters)of the particle from 0 is given by:-x=40+12t−t3. How long would the particles travel before coming to rest?

A person moves 30 m North and then 20 m towards East and finally 302​ m in South-West direction. The displacement of the person from the origin will be

The numerical ratio of displacement to distance for a moving object is

A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

Distinguish between Distance and displacement.

Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, then the number of electrons missing from each ion will be (e being the charge on an electron) :−

If the distance between earth and the sun were half of its present value , then how many number of days will be there in at year?

A farmer moves along the boundary of a square field of side $10$ m in $40$ sec What will be magnitude of displacement of the farmer at the end of $2$ minutes $20$ seconds from his initial position?

A particle moves along a straight line OX. At a time t(in seconds) the distance x(in meters)of the particle from 0 is given by:- $x = 40 + 12 t - t^{3}$ . How long would the particles travel before coming to rest?

A person moves $30 m$ North and then $20 m$ towards East and finally $302$ m in South-West direction. The displacement of the person from the origin will be

Two positive ions $,$ each carrying a charge $q,$ are separated by a distance $d .$ If $F$ is the force of repulsion between the ions $,$ then the number of electrons missing from each ion will be (e being the charge on an electron) $: -$

If the distance between earth and the sun were
half of its present value , then how many number of days will be there in at year?