Speed, average velocity and instantaneous velocity
Velocity
Velocity is the rate of change of displacement.
Instantaneous velocity
Instantaneous velocity is the velocity of a body at a specific instant in time.
Average velocity
Average velocity is the total displacement of a body over a time interval.
Velocity is the rate of change of position. It tells us how much an object's position changes in time. This is the same as the displacement divided by the time taken. Since displacement is a vector and time taken is a scalar, velocity is also a vector. We use the symbol
$v$ for velocity. If we have a displacement of
$\Delta x$ and a time taken of
$\Delta t$ ,
$v$ is then defined as:
Velocity can be positive or negative. Positive values of velocity mean that the object is moving away from the reference point or origin and negative values mean that the object is moving towards the reference point or origin.
An instant in time is different from the time taken or the time interval. It is therefore useful to use the symbol
$t$ for an instant in time (for example during the 4
^{th} second) and the symbol
$\Delta t$ for the time taken (for example during the first 5 seconds of the motion).
Average velocity (symbol
$v$ ) is the displacement for the whole motion divided by the time taken for the whole motion. Instantaneous velocity is the velocity at a specific instant in time.
(Average) Speed (symbol
$s$ ) is the distance travelled (
$d$ ) divided by the time taken (
$\Delta t$ ) for the journey. Distance and time are scalars and therefore speed will also be a scalar. Speed is calculated as follows:
Instantaneous speed is the magnitude of instantaneous velocity. It has the same value, but no direction.
James walks 2 km away from home in 30 minutes. He then turns around and walks back home along the same path, also in 30 minutes. Calculate James' average speed and average velocity.
The question explicitly gives
the distance and time out (2 km in 30 minutes)
the distance and time back (2 km in 30 minutes)
The information is not in SI units and must therefore be converted.
A man runs around a circular track of radius
$100\phantom{\rule{2pt}{0ex}}m$ . It takes him
$120\phantom{\rule{2pt}{0ex}}s$ to complete a revolution of the track. If he runs at constant speed, calculate:
his speed,
his instantaneous velocity at point A,
his instantaneous velocity at point B,
his average velocity between points A and B,
his average speed during a revolution.
his average velocity during a revolution.
To determine the man's speed we need to know the distance he travels and
how long it takes. We know it takes 120 s to complete one revolution ofthe track.(A revolution is to go around the track once.)
What distance is one revolution of the track? We know the
track is a circle and we know its radius, so we can determinethe distance around the circle. We start with
the equation for the circumference of a circle
Therefore, the distance the man covers in one revolution is
$\mathrm{628,32}\phantom{\rule{2pt}{0ex}}m$ .
We know that speed is distance covered per unit time. So if we divide the distance covered by the time it took we will know how much distance was covered for every unit of time. No direction is used here because speed is a scalar.
Consider the point A in the diagram.
We know which way the man is running around the track and we know hisspeed. His velocity at point A will be his speed (the magnitude of the
velocity) plus his direction of motion (the direction of hisvelocity).
The instant that he arrives at A he is moving as indicated in thediagram.
His velocity will be
$\mathrm{5,24}\phantom{\rule{2pt}{0ex}}m\xb7s{}^{-1}$ West.
Consider the point B in the diagram.
We know which way the man is running around the track and we know hisspeed. His velocity at point B will be his speed (the magnitude of the
velocity) plus his direction of motion (the direction of hisvelocity). The instant that he arrives at B he is moving as indicated in
the diagram.His velocity will be
$\mathrm{5,24}\phantom{\rule{2pt}{0ex}}m\xb7s{}^{-1}$ South.
To determine the average velocity between A and B, we need the change in
displacement between A and B and the change in time between A and B. Thedisplacement from A and B can be calculated by using the Theorem of Pythagoras:
Triangle AOB is isosceles and therefore angle BAO = 45
${}^{\circ}$ .
The direction is between west and south and is therefore southwest.
The final answer is:
$v=4.71\phantom{\rule{2pt}{0ex}}m\xb7s{}^{-1}$ , southwest.
Because he runs at a constant rate, we know that his speed anywhere around
the track will be the same. His average speed is
$\mathrm{5,24}\phantom{\rule{2pt}{0ex}}m\xb7s{}^{-1}$ .
Remember - displacement can be zero even when distance
travelled is not!
To calculate average velocity we need his total displacement and his total
time. His displacement is zero because he ends up where he started. Histime is
$120\phantom{\rule{2pt}{0ex}}s$ .
Using these we can calculate his average velocity:
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.