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\$

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>>Class 7
>>Physics
>>Motion and Time
>>Speed
>> $Speed-Basic$

Speed-Basic

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

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

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

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

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: 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: 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.

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) } } \$

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\$

Physics

definition

Speed

Speed is rate of change of distance,it is the measurement of how fast or slow objects move, can be given by formula,

Speed = \frac{Distance}{Time}

S.I unit of speed is

m / s

shortcut

Speed by distance moved and time taken by an object

Example : During inter school running race competition , Shivam completed the race of 200 m in 120 seconds, where as the winner of the race Anubhav completed it in just 80 seconds.What would be the required speed of Shivam to beat Anubhav?
Solution: Speed of Shivam

s_{1} = \frac{d i s t a n c e}{t i m e} = \frac{2 0 0}{1 2 0} = 1.67 m / s

Speed of Anubhav

s_{2} = \frac{d i s t a n c e}{t i m e} = \frac{2 0 0}{8 0} = 2.5 m / s

Hence to win the race speed of shivam should have been more than

2.5 m / s

REVISE WITH CONCEPTS

Uniform and Non- uniform Speed

ExampleDefinitionsFormulaes

LEARN WITH VIDEOS

Speed

8 mins

Quick Summary With Stories

Basic Introduction of Speed

2 mins

Speed and its Unit

2 mins

Important Questions

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:

Medium

View solution

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

Medium

View solution

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

Medium

View solution

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

Medium

View solution

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

Easy

View solution

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

Medium

View solution

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.

Easy

View solution

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

Medium

View solution

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.

Hard

View solution

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

Easy

View solution

definition

Speed

shortcut

Speed by distance moved and time taken by an object

REVISE WITH CONCEPTS

Uniform and Non- uniform Speed

Quick Summary With Stories

Basic Introduction of Speed

Speed and its Unit

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?

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.

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

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 $k m / h$

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.

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