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: 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 circle track of radius \$R\$ in \$40\$ seconds. What will be the displacement at the end of \$2\$ minutes \$20\$ seconds:

Total time duration in consideration \$=2\ min\ 20\ sec=140\ sec\$He completed 1 round in 40 secIn 140 sec, he will cover \$=\dfrac{140}{40}=3.5\ rounds\$After 3.5 rounds, he would be on the opposite side of the starting pointSo, the displacement will be the diameter or \$2R\$.

Q: 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\$

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: A woman starts from her home at 9.00 am, walks with a speed of \$5\, km\, h^{-1}\$ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of \$25\, km\, h^{-1}\$. Plot the position-time graph of the woman taking her home as origin.

Speed of the woman = \$5 \ km/h\$Distance between her office and home = \$2.5\$ kmTime taken = Distance / Speed= \$2.5 / 5 = 0.5\ h = 30 \ min\$It is given that she covers the same distance in the evening by an auto.Now, speed of the auto =\$ 25 \ km/h\$Time taken = Distance / Speed = \$2.5 / 25 = 1 / 10 = 0.1\ h = 6\ min\$The suitable x-t graph of the motion of the woman is shown in the given figure.

Question 1

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

Accepted Answer

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.

Question 2

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?

Accepted Answer

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.

Question 3

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?

Accepted Answer

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.

Question 4

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?

Accepted Answer

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=$10\sqrt{2}=14.14 m$

Question 5

An athlete completes one round of a circle track of radius $R$ in $40$ seconds. What will be the displacement at the end of $2$ minutes $20$ seconds:

Accepted Answer

Total time duration in consideration $=2\ min\ 20\ sec=140\ sec$He completed 1 round in 40 secIn 140 sec, he will cover $=\dfrac{140}{40}=3.5\ rounds$After 3.5 rounds, he would be on the opposite side of the starting pointSo, the displacement will be the diameter or $2R$.

Question 6

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

Accepted Answer

The distance travelled is given by $x=(t-2)^2 = t^2&#8211;4t + 4$As velocity is given by  $v=\dfrac{dx}{dt}$$v = 2t - 4$$Distance \ covered =\int|v|dt$=$ |t^2&#8211;4t|$$= 4t - t^2 for\ t = 0 to 2$=$ t^2 - 4t for \ t = 2 to 4$$= 8 units$

Question 7

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) $:-$

Accepted Answer

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

Question 8

A particle starts from the origin, goes along the $X-$axis to the point $(20\  m, 0)$ and then returns along the same line to the point $(-20\ m,0)$. Find the distance and displacement of the particle during the trip.

Accepted Answer

The starting point for the particle is origin as shown in figure.The Distance traveled by the particle is:$D =20+20+20$$=60 m$The displacement is the distance between the initial and the final points.Displacement $\Rightarrow OB=20 m$ in the negative directionThe distance is $60\ m$ and displacement is $20\ m$ in negative direction.

Question 9

A woman starts from her home at 9.00 am, walks with a speed of $5\, km\, h^{-1}$ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of  $25\, km\, h^{-1}$. Plot the position-time graph of the woman taking her home as origin.

Accepted Answer

Speed of the woman = $5 \ km/h$Distance between her office and home = $2.5$ kmTime taken = Distance / Speed= $2.5 / 5 = 0.5\  h = 30 \ min$It is given that she covers the same distance in the evening by an auto.Now, speed of the auto =$ 25 \ km/h$Time taken = Distance / Speed = $2.5 / 25 = 1 / 10 = 0.1\  h = 6\  min$The suitable x-t graph of the motion of the woman is shown in the given figure.

Question 10

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Accepted Answer

time    step  distance      0  0  0    1  1  1    2  2  2    3  3  3    4  4  4    5  5  5    6  6  4    7  7  3    8  8  2    9  9  3    10  10  4    11  11  5    12  12  6    13  13  7    14  14  6    15  15  5    16  16  4    17  17  5    18  18  6    19  19  7    20  20  8    21  21  9    22  22  8    23  23  7    24  24  6    25  25  7    26  26  8    27  27  9    28  28  10    29  29  11    30  30  10    31  31  9    32  32  8    33  33  9    34  34  10    35  35  11    36  36  12    37  37  13 Lets call 1 cycle = going forward + coming backIn 1 cycle, he goes 2 steps = $2 \ m$ forwardIn 1 cycle, he spends time of 8 unitsSo, in 4 cycles, displacement $= 8 \ m$, time = 32 unitsFor 5th cycle, man reaches pit in 5 steps and falls in the pitSo, total time = $32 + 5 = 37$ units

Uniform and Non-Uniform Motion

Language

Rate

REVISE WITH CONCEPTS

Introduction to Motion

Understanding Uniform and Non-Uniform Motion

Uniform and Non- uniform Speed

Uniform And Non-Uniform Velocity

QUICK SUMMARY WITH STORIES

Basics of Motion

Distance and Displacement

Understanding Uniform Motion

Understanding Non-Uniform Motion

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

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

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?

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?

An athlete completes one round of a circle track of radius $R$ in $40$ seconds. What will be the displacement at the end of $2$ minutes $20$ seconds:

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

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) $: -$

A particle starts from the origin, goes along the $X -$ axis to the point $(20 m, 0)$ and then returns along the same line to the point $(- 20 m, 0)$ . Find the distance and displacement of the particle during the trip.

A woman starts from her home at 9.00 am, walks with a speed of $5 k m h^{- 1}$ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of $25 k m h^{- 1}$ . Plot the position-time graph of the woman taking her home as origin.

Uniform and Non-Uniform Motion

Language

Rate

REVISE WITH CONCEPTS

Introduction to Motion

Understanding Uniform and Non-Uniform Motion

Uniform and Non- uniform Speed

Uniform And Non-Uniform Velocity

QUICK SUMMARY WITH STORIES

Basics of Motion

Distance and Displacement

Understanding Uniform Motion

Understanding Non-Uniform Motion

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

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

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?

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?

An athlete completes one round of a circle track of radius R in 40 seconds. What will be the displacement at the end of 2 minutes 20 seconds:

The displacement of a particle is given by x=(t−2)2 where x is in metres and t is in seconds. The distance covered by the particle in first 4 seconds is

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) :−

A particle starts from the origin, goes along the X−axis to the point (20 m,0) and then returns along the same line to the point (−20 m,0). Find the distance and displacement of the particle during the trip.

A woman starts from her home at 9.00 am, walks with a speed of 5kmh−1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25kmh−1. Plot the position-time graph of the woman taking her home as origin.

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

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 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?

An athlete completes one round of a circle track of radius $R$ in $40$ seconds. What will be the displacement at the end of $2$ minutes $20$ seconds:

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

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) $: -$

A particle starts from the origin, goes along the $X -$ axis to the point $(20 m, 0)$ and then returns along the same line to the point $(- 20 m, 0)$ . Find the distance and displacement of the particle during the trip.

A woman starts from her home at 9.00 am, walks with a speed of $5 k m h^{- 1}$ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of $25 k m h^{- 1}$ . Plot the position-time graph of the woman taking her home as origin.