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definition of non uniform motion
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Some examples of nonuniform circular motion include a roller coaster, a vertical pendulum, .... This page was last modified on 5 November 2010 at 05:57. ...
Linear motion is motion along a straight line, and can therefore be described mathematically using only one spatial dimension. It can be uniform, that is, with constant velocity (zero acceleration), or nonuniform, that is, with a variable velocity (nonzero acceleration). The motion of a particle (a pointlike object) along the line can be described by its position x, which varies with t (time). Linear motion is sometimes called rectilinear motion.
An example of linear motion is that of a ball thrown straight up and falling back straight down.
The average velocity v during a finite time span of a particle undergoing linear motion is equal to
 v = \frac {\Delta d}{\Delta t}.
The instantaneous velocity of a particle in linear motion may be found by differentiating the position x with respect to the time variable t. The acceleration may be found by differentiating the velocity. By the fundamental theorem of calculus the converse is also true: to find the velocity when given the acceleration, simply integrate the acceleration with respect to time; to find displacement, simply integrate the velocity with respect to time.
This can be demonstrated graphically. The gradient of a line on the displacement time graph represents the velocity. The gradient of the velocity time graph gives the acceleration while the area under the velocity time graph gives the displacement. The area under an acceleration time graph gives the velocity.
Linear motion is the most basic of all motions. According to Newton's first law of motion, objects not subjected to forces will continue to move uniformly in a straight line indefinitely. Under everyday circumstances, external forces such as gravity and friction will cause objects to deviate from linear motion and can cause them to come to a rest.
For linear motion embedded in a higherdimensional space, the velocity and acceleration should be described as vectors, made up of two parts: magnitude and direction. The direction part of these vectors is the same and is constant for linear motion, and only for linear motion
In mechanics, the derivative of the position vs. timegraph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the yaxis and time on the xaxis, the slope of the curve is given by:
 v = \frac{\Delta y}{\Delta x} = \frac{\Delta s}{\Delta t}.
Here s is the position of the object, and t is the time. Therefore, the slope of the curve gives the change in position (in metres) divided by the change in time (in seconds), which is the definition of the average velocity (in meters per second (\begin{matrix} \frac{m}{s} \end{matrix})) for that interval of time on the graph. If this interval is made to be infinitesimally small, such that {\Delta s} becomes {ds} and {\Delta t} becomes {dt}, the result is the instantaneous velocity at time t, or the derivative of the position with respect to time.
A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the yaxis and time on the xaxis. Again the slope of a line is change in y over change in x:
 a = \frac{\Delta y}{\Delta x} = \frac{\Delta v}{\Delta t}.
Where v is the velocity, measured in \begin{matrix} \frac{m}{s} \end{matrix}, and t is the time measured in seconds. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives \begin{matrix} \frac{dv}{dt} \end{matrix}, the instantaneous acceleration at time t, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). The units of this slope or derivative are in meters per second per second (\begin{matrix} \frac{m}{s^2} \end{matrix}, usually termed "meters per secondsquared"), and so, therefore, is the acceleration.
Since the acceleration of the object is the second derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the yaxis and time on the xaxis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. \begin{matrix} \frac{m}{s} \end{matrix}s = m.)
The same multiplication rule holds true for acceleration vs. time graphs. When (\begin{matrix} \frac{m}{s^2} \end{matrix}) is multiplied by time (s), velocity is obtained. (\begin{matrix} \frac{m}{s^2} \end{matrix}s = \begin{matrix} \frac{m}{s} \end{matrix}).
Variable rates of change
The expressions given above apply only when the rate of change is constant or when only the average (mean) rate of change is required. If the velocity or positions change nonlinearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the timespans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are
 v = \frac{ds}{dt},
 a = \frac{dv}{dt}.
Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to position:
 a = \frac{d^2 s}{dt^2}.
Since, for the purposes of mechanics such as this, integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:
 s(t_2)s(t_1) = \int_{t_1}^{t_2}{v}\, dt,
 v(t_2)v(t_1) = \int_{t_1}^{t_2}{a}\, dt.
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Answers:uniform motion is not necisarily motion atthe same speed. uniform motion is where there is some regularity in the motion. an equal and exponential increase in velocity, is considered uniform motion.
Answers:Q1. Uniform motion definition (my guess): Same distance covered every second, which also means a constant velocity. Free fall has an increasing velocity every second. Q2. Yes. You are accelerating right now while reading this. Gravity is causing you to ALWAYS accelerate while on earth. Q3. Simple harmonic motion Q4. Angular acceleration Q5. The displacement is 0 (starting position minus ending position).The distance covered is 2*pi*r (circumference)
Answers:Let the distance be x Speed = 15kmph Time = x/15 hours Let the time he has to reach be y hours x/15 = y  1 y = x/15 + 1 (A) x/ 10 = y + 1 y = x/10  1 (B) (A) = (B) x/15 + 1 = x/10 1 x/10  x/15 = 2 MULTIPLY BY 30 3x  2x = 60 x = 60 km 60/15 = 4 hours 60/5 = 12 km per hour ANSWER CHECK 60/15 = 4 hours 60/10 = 6 hours 60/12 = 5 hours
Answers:Motion implies momentum, which implies velocity. Linear implies a straight line. Accelerating implies changing velocity. And uniform implies constancy. So, when a body moves in a straight line and accelerates at a constant rate, the body is said to have an uniformly accelerated linear motion.
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