Q: An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:\$\frac{dv}{dt} = -2.5\sqrt{v}\$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

\$According\, \, to\, question................... \\ \int _{ 6.25 }^{ 0 }{ \left( { \dfrac { { dv } }{ { \sqrt { v } } } } \right) =\, -2.5\int _{ 0 }^{ t }{ dt } } \\ \Rightarrow \left| { 2\sqrt { v } } \right| _{ 6.25 }^{ 0 }=-2.5\, t \\ \Rightarrow 2.\, \sqrt { 6.25 } =2.5\, t \\ \, \, \therefore \, \, \, t=2\, { { s } } \\ so\, the\, correct\, optuon\, is\, A\$

Q: A body travelling with uniform acceleration crosses two points \$A\$ and \$B\$ with velocities \$20\ m/sec\$ and \$30\ m/sec\$ respectively then the speed of the body at mid point of \$A\$ and \$B\$ is

Let the acceleration of the car is \$= a\$ and the distance between \$A \$ and \$B\$ is \$=d\$\$v^2-u^2=2ad\$Given,\$v=30\,m/s\$ and \$u=20\,m/s\$\$2ad=30^2-20^2\$\$ad=\dfrac{(900-400)}{2}=250\$when the car is at the mid point of \$AB\$ then let the speed of the car is \$v_1\$\$v_1^2-20^2=2a(\dfrac d2)\$\$v_1^2=ad+400=250+400=650\$therefore \$v_1\$ is \$= 25.5\,m/s\$

Q: Change the speed of \$6\ m/s\$ into \$km/h\$

Given that, Speed \$v=6\,m/s\$ Now, speed in \$km/h\$ \$ v=6\times \dfrac{18}{5} \$ \$ v=21.6\,km/h \$ Hence, the speed is \$21.6\ km/h\$

Q: Two electrons lying \$10 cm\$ apart are released. What will be their speed when they are \$20 cm\$ apart?

Let their velocities at 20cm a part be v. Then \$ \Delta KE_{Total} = 2[\dfrac{Me}{2}v^{2}-\dfrac{m}{2}(0)^{2}] = Mev^{2} \$ Potential energy 10 cm (U Initial ) \$ = \dfrac{-1ke^{2}}{\dfrac{10}{100}} = 10ke^{2} \$Potential Energy 20 cm (U final ) \$ = \dfrac{ke^{2}}{20/100} = 5ke^{2} \$\$ \Delta U_{Total} = U_{final }-U_{initial} \$\$ = 5ke^{2}-10ke^{2} = -5ke^{2} \$As electrostatics force is conservative, \$ \Delta KE _{Total}+\Delta PE_{Total} = 0 \$\$ \Rightarrow mev^{2}-5ke^{2} = 0 \$\$ \Rightarrow v = \sqrt{\dfrac{5ke^{2}}{me}} = 35.58m/s \$

Q: The distance between two stations is \$240\$km. A train takes \$4\$ hours to cover this distance. Calculate the speed of the train.

Distance \$=240\$kmTime taken \$=4\$ hoursSpeed\$=\$ Distance covered \$/\$time taken \$=240km/4\$ hr \$=60km/h\$Speed of train \$=60km/h\$.

Q: The odometer of a car reads \$57321.0\$km when the clock shows the time \$08:30\$AM. What is the distance moved by the car, if at \$08:50\$ AM, the odometer reading has changed to \$57336.0\$km \$?\$ Calculate the speed of the car in km\$/\$min during this time. Express the speed in km\$/\$h also.

Distance \$= 57336.0km - 57321km = 15km\$Speed in km\$/\$min \$=15km/20min = \dfrac{3}{4} km/min\$Speed in \$km/hr= \dfrac{3}{4} \times \dfrac{km}{min} \times \dfrac{60 min }{hr} =(15\times 3)km/hr =45km/hr\$.

Q: The distance between two stations is 240 km. A train takes 4 hrs. to cover this distance. Calculate the speed of the train.

Distance covered by train \$S = 240 \ km\$Time taken by train to cover this distance \$t = 4 \ hrs\$Thus speed of train \$v = \dfrac{S}{t}\$\$\therefore \ v = \dfrac{240}{4} = 60 \ km/hr\$

Question 1

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{d v}{d t} = - 2.5 v$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

Accepted Answer

$According\, \, to\, question................... \ \int  _{ 6.25 }^{ 0 }{ \left( { \dfrac { { dv } }{ { \sqrt { v }  } }  } \right) =\, -2.5\int  _{ 0 }^{ t }{ dt } } \ \Rightarrow \left| { 2\sqrt { v }  } \right| _{ 6.25 }^{ 0 }=-2.5\, t \ \Rightarrow 2.\, \sqrt { 6.25 } =2.5\, t \ \, \, \therefore \, \, \, t=2\, { { s } } \ so\, the\, correct\, optuon\, is\, A$

Question 2

A bus travels 54 km in 90 minutes. The speed of the bus is?

Accepted Answer

speed = distance / time           =54 * 1000 /90 * 60           =10 m/s

Question 3

A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20\ m/sec$ and $30\ m/sec$ respectively then the speed of the body at mid point of $A$ and $B$ is

Accepted Answer

Let the acceleration of the car is $= a$  and the distance between $A $ and $B$  is $=d$$v^2-u^2=2ad$Given,$v=30\,m/s$ and $u=20\,m/s$$2ad=30^2-20^2$$ad=\dfrac{(900-400)}{2}=250$when the car is at the mid point of $AB$ then let the speed of the car is $v_1$$v_1^2-20^2=2a(\dfrac d2)$$v_1^2=ad+400=250+400=650$therefore $v_1$ is $= 25.5\,m/s$

Question 4

Change the speed of $6\ m/s$ into $km/h$

Accepted Answer

Given that,&#10;&#10;Speed $v=6\,m/s$ &#10;&#10;Now, speed in $km/h$&#10;&#10;  $ v=6\times \dfrac{18}{5}&#10;$ &#10;&#10; $ v=21.6\,km/h $ &#10;&#10;Hence, the speed is $21.6\ km/h$

Question 5

What does the path of an object look like when it is in uniform motion?

Accepted Answer

In uniform motion, the object moves along a straight path i.e. the path of object is a straight line.

Question 6

Two electrons lying $10 cm$ apart are released. What will be their speed when they are $20 cm$ apart?

Accepted Answer

Let their velocities at 20cm a part be v. Then $ \Delta KE_{Total} = 2[\dfrac{Me}{2}v^{2}-\dfrac{m}{2}(0)^{2}] = Mev^{2} $ Potential energy 10 cm (U Initial ) $ = \dfrac{-1ke^{2}}{\dfrac{10}{100}} = 10ke^{2} $Potential Energy 20 cm (U final ) $ = \dfrac{ke^{2}}{20/100} = 5ke^{2} $$ \Delta U_{Total} = U_{final }-U_{initial} $$ = 5ke^{2}-10ke^{2} = -5ke^{2} $As electrostatics force is conservative, $ \Delta KE _{Total}+\Delta PE_{Total} = 0 $$ \Rightarrow mev^{2}-5ke^{2} = 0 $$ \Rightarrow v = \sqrt{\dfrac{5ke^{2}}{me}} = 35.58m/s $

Question 7

The distance between two stations is $240$km. A train takes $4$ hours to cover this distance. Calculate the speed of the train.

Accepted Answer

Distance $=240$kmTime taken $=4$ hoursSpeed$=$ Distance covered $/$time taken $=240km/4$ hr                         $=60km/h$Speed of train $=60km/h$.

Question 8

The odometer of a car reads $57321.0$km when the clock shows the time $08:30$AM. What is the distance moved by the car, if at $08:50$ AM, the odometer reading has changed to $57336.0$km $?$ Calculate the speed of the car in km$/$min during this time. Express the speed in km$/$h also.

Accepted Answer

Distance $= 57336.0km - 57321km = 15km$Speed in km$/$min $=15km/20min = \dfrac{3}{4} km/min$Speed in $km/hr= \dfrac{3}{4} \times \dfrac{km}{min} \times \dfrac{60 min }{hr} =(15\times 3)km/hr =45km/hr$.

Question 9

A uniform spherical shell of mass $M$ and radius $R$ rotates about a vertical axis on frictionless bearing. A massless cord passes around the equator of the shell, over a pulley of rotational inertia $I$ and radius $R$ and is attached to a small object of mass $m$ that is otherwise free to fall under the influence of gravity. There is no friction of pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen a distance $h$ from rest? Use work-energy considerations.

Accepted Answer

By conservation of energy, when object falls by a distance of h.$\cfrac { 1 }{ 2 } m{ v }^{ 2 }+\cfrac { 1 }{ 2 } { I }_{ shell }{ \omega  }_{ shell }^{ 2 }+\cfrac { 1 }{ 2 } { I }_{ pulley }=mgh$$v=R\omega $$\cfrac { 1 }{ 2 } m{ v }^{ 2 }+\cfrac { 1 }{ 2 } { \left( \cfrac { 2 }{ 3 } M{ R }^{ 2 } \right)  }\times { \left( \cfrac { v }{ R }  \right)  }^{ 2 }+\cfrac { 1 }{ 2 } { I }_{ p }{ \left( \cfrac { v }{ { r }_{ p } }  \right)  }^{ 2 }=mgh$${ v }^{ 2 }\left( m+\cfrac { 2 }{ 3 } M+\cfrac { { I }_{ p } }{ { r }_{ p }^{ 2 } }  \right) =2mgh$$v=\sqrt { \cfrac { 2mgh }{ \left( m+\cfrac { 2 }{ 3 } M+\cfrac { { I }_{ p } }{ { r }_{ p }^{ 2 } }  \right)  }  } $

Question 10

The distance between two stations is 240 km. A train takes 4 hrs. to cover this distance. Calculate the speed of the train.

Accepted Answer

Distance covered by train   $S = 240 \ km$Time taken by train to cover this distance   $t = 4 \ hrs$Thus speed of train   $v = \dfrac{S}{t}$$\therefore \ v = \dfrac{240}{4} = 60 \ km/hr$

Revise with Concepts

Speed-Basic

Uniform and Non- uniform Speed

Quick Summary With Stories

Basic Introduction of Speed

Speed and its Unit

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

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{d v}{d t} = - 2.5 v$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

A bus travels 54 km in 90 minutes. The speed of the bus is?

A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 m / s e c$ and $30 m / s e c$ respectively then the speed of the body at mid point of $A$ and $B$ is

Change the speed of $6 m / s$ into $k m / h$

What does the path of an object look like when it is in uniform motion?

Two electrons lying $10 c m$ apart are released. What will be their speed when they are $20 c m$ apart?

The distance between two stations is $240$ km. A train takes $4$ hours to cover this distance. Calculate the speed of the train.

The odometer of a car reads $57321.0$ km when the clock shows the time $08 : 30$ AM. What is the distance moved by the car, if at $08 : 50$ AM, the odometer reading has changed to $57336.0$ km $?$ Calculate the speed of the car in km $/$ min during this time. Express the speed in km $/$ h also.

The distance between two stations is 240 km. A train takes 4 hrs. to cover this distance. Calculate the speed of the train.

Revise with Concepts

Speed-Basic

Uniform and Non- uniform Speed

Quick Summary With Stories

Basic Introduction of Speed

Speed and its Unit

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

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by: dtdv​=−2.5v​ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

A bus travels 54 km in 90 minutes. The speed of the bus is?

A body travelling with uniform acceleration crosses two points A and B with velocities 20 m/sec and 30 m/sec respectively then the speed of the body at mid point of A and B is

Change the speed of 6 m/s into km/h

What does the path of an object look like when it is in uniform motion?

Two electrons lying 10cm apart are released. What will be their speed when they are 20cm apart?

The distance between two stations is 240km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

The odometer of a car reads 57321.0km when the clock shows the time 08:30AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0km ? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.

The distance between two stations is 240 km. A train takes 4 hrs. to cover this distance. Calculate the speed of the train.

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
$\frac{d v}{d t} = - 2.5 v$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 m / s e c$ and $30 m / s e c$ respectively then the speed of the body at mid point of $A$ and $B$ is

Change the speed of $6 m / s$ into $k m / h$

Two electrons lying $10 c m$ apart are released. What will be their speed when they are $20 c m$ apart?

The distance between two stations is $240$ km. A train takes $4$ hours to cover this distance. Calculate the speed of the train.

The odometer of a car reads $57321.0$ km when the clock shows the time $08 : 30$ AM. What is the distance moved by the car, if at $08 : 50$ AM, the odometer reading has changed to $57336.0$ km $?$ Calculate the speed of the car in km $/$ min during this time. Express the speed in km $/$ h also.