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kinematic parameters

kinematic parameters
Category : Kinematic

Learn about kinematic's topics in this section with a problem-based and concise way.


kinematic parameters

kinematic parameters

Kinematic

Mechanics is one of the main subjects in any physics courses of high schools and colleges. The concept of Kinematic, which is a topic of electrostatic, are taught in a simple, elegant and concise way in this section with some typical problems with descriptive answer. You can solve more problems in the ExamCenter to strengthen your understanding and get better grade in your final or midterm exams.

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Kinematic is a branch of classical mechanics which describes the motion of point particles, systems of distributed mass and other forms of objects without consideration of the causes of the motion in terms of position, velocity and acceleration.


Displacement

We want to study the car’s motion along a straight line. Let the car to be a point particle. To specify the location of the particle in one dimension we need only one axis that we called it $x$ and lies along the straight line path. First we must define an important quantity that the other quantities are made from it, Displacement. To describe the motion of a particle, we must know its position and how that position changes with time. The change in the particle’s position from initial position $x_i$ to final position $x_f$ is called displacement, $x_f-x_i$. In physics we use the Greek letter $\Delta$ to indicate the change in a quantity. So the change in position, $x$, or displacement can be written as $\Delta x=x_f-x_i$. In the figure below, the particle moves from point $A$ at $x=2\, {\rm m}$ and after reaching to $x=9\,{\rm m}$ returns and stops at position $x=6\,{\rm m}$ at point $B$. The displacement of the particle is a vector points from $A$ to $B$ and denoted by $\Delta x=x_B-x_A$. Displacement is a vector that depends only on initial and final positions of the particle and not on the details of the motion and path. it requires both a length and a direction to be described. Therefore, the car’s displacement is $\Delta x=6-2=+4\,{\rm m}$. the positive sign indicates the particle finally displaced toward the positive $x$ direction. The distance traveled or the overall path the particle has covered is a scalar quantity which in this case is equal to $10\,{\rm m}$. in general, the distance traveled and the magnitude of the displacement vector between two points are not the same.


Average velocity and speed:

We usually need to know at what rate does the particle moves? To answer this question, we define an important quantity  as follows: \[ Average\ speed=\frac{distance\ traveled}{elapsed\ time}=\frac{s}{\Delta t}\]
\[Average\ velocity=\frac{displacement}{elapsed\ time}\]

The average speed defines in terms of the distance traveled so it is a scalar quantity. But the average velocity is a vector that points in the same direction as the displacement vector. The average velocity is a more useful quantity than the other since it describes both how fast and in what direction the objects move. Therefore, in one direction, say $x$,the average velocity is defined as \[v_{ave-x}=\frac{\Delta x}{\Delta t}\]

The SI unit of velocity is meters per second ($\rm m/s$). Other common units include Kilometers per second ($\rm Km/h$), feet per second ($\rm ft/s$) and miles per hour ($\rm mi/h$).

Note that, in general, the magnitude of the average velocity and average speed are not equal in a given time interval. For example, if the above car moves from $A$ to $B$ in $4\,{\rm s}$ we have \[Average\ speed = \frac{10\,{\rm m}}{4\,{\rm s}}=2.5\ {\rm m/s}\] whereas \[Average\ velocity= \frac{\Delta x}{\Delta t}=+\frac{4\,{\rm m}}{4\,{\rm s}}=+1\ {\rm m/s}\]
The positive indicates that the overall motion is in the positive $x$ direction.

Example: a bird is flying $100\,{\rm m}$ due east at $10\,{\rm m/s}$ and then it turns around and flying west in $15\,{\rm s}$ at $20\,{\rm m/s}$. Find the average velocity and average speed during the overall time interval.

Solution:
First we must find the overall time. The first and second parts are done in $\Delta t_1=\frac{\Delta x}{v}=\frac{100}{10}=10\,{\rm s}$ and $\Delta t_2=15\, {\rm s}$. Thus the overall time of flying of the bird is $\Delta t_{tot}=10+15=25\,{\rm s}$. the traveled distance of the bird due west is $d_w=v\Delta t_2=20 \times 15=300\,{\rm m}$. therefore, by definition of the average speed, we have
\[Average\ speed =\frac{(100+300)\,{\rm m}}{25\,{\rm s}}=16\ {\rm m/s}\]
As we can see from the figure, the displacement vector is $\Delta x=x_f-x_i=-200-0=-200\,{\rm m}$. so the average velocity is found as
\[v_{ave}=\frac{\Delta x}{\Delta t}=\frac{-200\,{\rm m}}{25\,{\rm s}}=-8\ {\rm m/s}\]
The negative indicates that the average velocity is towards the $–x$ direction.

To show the motion of a particle is usually used from a position–versus-time ($x-t$) graph. If the velocity of the motion is constant during any successive equal-time intervals, then the $x-t$ graph is a straight-line. This type of motion is called uniform motion. In this case, the average velocity is the slope of the position-versus-time graph. In one dimensional motion this is simply
\[v_{ave-x}=\frac{\Delta x}{\Delta t}={\rm slope\ of\ the\ x-t\ graph}\]
Therefore, from the mathematical viewpoint, in the uniform motion equal displacements occur during any successive equal-time interval.
If the velocity is not constant, the motion’s graph is a curve as shown in the figure below. In such a case, the object’s average velocity during each particular time interval is the slope of the $x-t$ graph that connects the initial and final points in that interval $\Delta t$.

Instantaneous velocity

The average velocity is not a useful quantity for analyzing the non-uniform motions. Very often it is necessary to know the object’s velocity at a single instant of time. In this case the instantaneous velocity can be defined. Instantaneous velocity is the limit of average velocity $v_{ave}=\frac{\Delta x}{\Delta t}$ as $\Delta t$ approaches zero. In the language of calculus, this limit is called the derivative of $x$ with respect to $t$ and is written $\frac{dx}{dt}$. Thus in the mathematical form we can state 
\[v_x=\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}\]
Therefore, the instantaneous velocity at time $t$ is the slope of the line that is tangent to the position-versus-time graph at the instance $t$. The time interval $\Delta t$ is always positive, so $v_x$ has the same sign as $\Delta x$.


Example: the position of a particle varies with time according to the equation $x=40-5t-5t^2$ where $x$ and $t$ are in $\rm m$ and $\rm s$, respectively. Find 
($a$)    The particle’s average velocity between $t=1\,{\rm s}$ and $t=2\,{\rm s}$. 
($b$)    The velocity at $t=2\,{\rm s}$.

Solution

        (a) By definition of the average velocity we must first find the displacement during that interval
                \[ x(t=2\,{\rm s})=40-5(2)-5(2)^2=10\ {\rm m} \]
                \[ x(t=1\,{\rm s})=40-5(1)-5(1)^2=30\ {\rm m} \]
                \[ \Delta x=x(t=2\,{\rm s})-x(t=1\,{\rm s})=10-30=-20\ {\rm m} \]
                Therefore, the average velocity becomes
                \[v_{ave-x}=\frac{\Delta x}{\Delta t}=\frac{-20\,{\rm m}}{(2-1)\,{\rm s}}=-20\ {\rm m/s}\]
        (b) The velocity is $v_x=\frac{dx}{dt}$. So 
                \[v_x=\frac{dx}{dt}=\frac{d}{dt} (40-5t-5t^2)=(-5-10\,t)\ {\rm m/s}\]
                Evaluating the velocity at $t=2\,{\rm s}$ gives
                \[v_x (t=2\,{\rm s})=-5-10(2)=-25\ {\rm m/s}\]
                In both cases the negatives indicate the particle moves in the $–x$ direction.


Acceleration

Acceleration means the rate of change of velocity, that is, acceleration measures how quickly or slowly an object’s velocity changes. this change may be the result of the change in the magnitude or direction of the velocity or both of them. Thus if the magnitude of the velocity remains constant and only its direction varies, we can still have an accelerating motion (for example in the circular motions).

The average acceleration during a particular time interval $\Delta t$ is defined as
\[a_{ave}=\frac{\Delta v}{\Delta t}\]

The SI unit of acceleration is (meters per second) per second, abbreviated $\rm m/s^2$. The average acceleration like other kinematic parameters position $r$ and velocity $v$ is a vector and points in the same direction as velocity change $\Delta v$.
If the velocity of a car changes from $0$ to $50\,{\rm km/h}$ in $10\,{\rm s}$, its average acceleration is found as
\[a_{ave}=\frac{50\,(\rm Km/h)}{10\,{\rm s}}\]
This means that the velocity of the car increases by $5\, {\rm km/h}$ every $1\,{\rm s}$.

Instantaneous acceleration is the limit of the ratio $\frac{\Delta v}{\Delta t}$ as $\Delta t$ approaches zero or in the mathematical language is the derivative of the velocity with respect to time. 
\[a=\lim_{\Delta t \to 0}\frac{\Delta v}{\Delta t}=\frac{dv}{dt}\]

The algebraic sign of the acceleration doesn’t tell us about the object is speeding up or slowing down. To determine this, we must take into account the signs of both of them. In one dimensional motion, say $x$ direction, if $v_x$ and $a_x$ are both positive, the speed of the object is increasing and it is speeding up. When $v_x$ and $a_x$ have opposite signs, the speed is again increasing. In summary, when $v_x$ and $a_x$ have the same sign, the speed is increasing and when they have opposite signs, the speed is decreasing which usually called decelerating motion.

 

Example: A particle at time $t=3\, {\rm s}$ and at position $x=7\, {\rm m}$ has velocity $v=4\, {\rm m/s}$. At $t=7\, {\rm s}$ it is located at $x=-5\, {\rm m}$ with velocity $v=-2\, {\rm m/s}$. Find:
(a)    The average velocity of the particle.
(b)    The average acceleration of the particl
e.

Solution

(a) By definition of the average velocity, we must first determine the displacement vector.so 
\[\Delta x=x_f-x_i=(-5\,{\rm m})-(7\, {\rm m})=-12\ {\rm m}\]
\[average\ velocity : v_{ave}=\frac{\Delta x}{\Delta t}=\frac{-12\, {\rm m}}{(7-3)\, {\rm s}}=\frac{-12}{4}=-3\ {\rm m/s}\]
The minus sign indicates that the direction of the motion of the particle points to the $–x$ axis.
(b) To find the average acceleration, first determine the change in the velocity of the particle during that interval time.
\[ \Delta v=v_f-v_i=-2-4=-6\ {\rm m/s}\]
By definition we have
\[ a_{ave}=\frac{\Delta v}{\Delta t}=\frac{-6\, {\rm m/s}}{(7-3)\, {\rm s}}=\frac{-6}{4}=-1.5\ {\rm m/s^2}\]


 

Displacement,Average Velocity,Average acceleration,Speed,Instantaneous velocity,Instantaneous acceleration,kinematic