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\\)

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

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

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\\)

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\\)

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.

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 \\)

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\\).

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\\).

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) } } \\)

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\\)

Uniform and Non- uniform Speed | Definition, Examples, Diagrams

definition

Uniform motion

A body is said to be in uniform motion if its velocity does not change with time. For a body in uniform motion, net external force is always equal to zero. An example of uniform motion is a body sliding on a frictionless surface with a constant velocity.

definition

Non-uniform Motion

A body is said to have non-uniform motion if its velocity changes with time. Such a body is necessarily under the influence of an external force. An example of non-uniform motion is a body rotating at a constant speed. Its velocity is not constant because its direction of motion changes every instant.

example

Difference between uniform and non-uniform motion

Uniform Motion	Non-Uniform Motion
When an object travels equal distance in equal interval of time .	When an object travels unequal distance in equal interval of time.
For example : If a car covers equal distance of 500m in the interval of 2 minutes.	For example : Motion of car on a crowded road when it covers unequal distance in equal interval of time.

Uniform and Non- uniform Speed

definition

Uniform motion

definition

Non-uniform Motion

example

Difference between uniform and non-uniform motion

REVISE WITH CONCEPTS

Speed-Basic

Quick Summary With Stories

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.

definition

Uniform motion

definition

Non-uniform Motion

example

Difference between uniform and non-uniform motion

REVISE WITH CONCEPTS

Speed-Basic

Quick Summary With Stories

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.