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.

The displacement of a particle is given by \$ x={ \left( t-2 \right) }^{ 2 } \$ where x is in metres and t is in seconds. The distance covered by the particle in first 4 seconds is

The distance travelled is given by \$x=(t-2)^2 = t^2–4t + 4\$As velocity is given by \$v=\dfrac{dx}{dt}\$\$v = 2t - 4\$\$Distance \ covered =\int|v|dt\$=\$ |t^2–4t|\$\$= 4t - t^2 for\ t = 0 to 2\$=\$ t^2 - 4t for \ t = 2 to 4\$\$= 8 units\$

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>>Class 11
>>Physics
>>Motion in a Straight Line
>>Distance and Displacement
>> $Coordinate System and Position Vector$

Coordinate System and Position Vector

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: 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 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: 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) \$:-\$

Two positive ions each carrying a charge \$q\$ are kept at a distance \$d\$,it is found that force of repulsion between them is\$\begin{array}{l} F=\dfrac { { kqq } }{ { { d^{ 2 } } } } \\ =\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } \dfrac { { qq } }{ { { d^{ 2 } } } } \\ where\, \, q=ne \\ F=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } \dfrac { { { n^{ 2 } }{ e^{ 2 } } } }{ { { d^{ 2 } } } } \\ n=\sqrt { \dfrac { { 4\pi { E_{ 0 } }F{ d^{ 2 } } } }{ { { e^{ 2 } } } } } \end{array}\$

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: 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: A body travels a distance of \$3\ km\$ towards East, then \$4\ km\$ towards North and finally \$9\ km\$ towards East. (i) What is the total distance travelled? (ii) What is the resultant displacement?

(i) Total distance travelled \$= 3 + 4 +9=16\ km\$(ii) The body travels a total distance of \$12\ km\$ in east direction i.e. towards x-axis. And it travels a distance of \$4\ km\$ in North direction, i.e. towards y-axis. Hence, resultant displacement is \$=\sqrt{12^2 + 4^2} = \sqrt{144+ 16} = \sqrt{160} = 12.6km\$

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