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# examples of rectilinear motion

From Wikipedia

Linear motion

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 non-uniform, that is, with a variable velocity (non-zero acceleration). The motion of a particle (a point-like 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 every-day 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 higher-dimensional 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

Projectile motion

Projectile motion is one of the traditional branches of classical mechanics, with applications to ballistics. A projectile is any body that is given an initial velocity and then follows a path determined by the effect of the gravitational acceleration and by air resistance. Projectile motion is the motion of such a projectile. For example, a thrown football, an object dropped from an airplane, or a bullet shot from a gun are all examples of projectiles. Projectile motion may only be used to solve mechanics problems if the acceleration is constant.

The path followed by a projectile is called its trajectory. It is affected by gravity.

The initial velocity v0 can be written as

v_0=v_{0x}\mathbf{i} + v_{0y}\mathbf{j}

The components v0x and v0y can be found if the angle Î¸0 is known:

v_{0x}=v_0\cos\theta_0

and

v_{0y}=v_0\sin\theta_0

## Without air resistance

The horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

Since there is no acceleration in the horizontal direction, the horizontal component of the velocity remains unchanged throughout the motion.

The horizontal displacement

x-x_0

from an initial position x0 is given by the equation

x-x_0=v_{0x}t+\tfrac{1}{2}at^2,

with a = 0 and v_{0x}=v_0\cos\theta_0. Thus

x-x_0=(v_0\cos\theta_0)t.

The vertical motion is the motion of a particle in free fall. Equations for free fall apply. For example, y-y_0=(v_0\sin\theta_0)t-\tfrac{1}{2}gt^2. Other useful equations for the vertical y-axis are v_y=v_0\sin\theta_0-gt, and v_y^2=(v_0\sin\theta_0)^2-2g(y-y_0).

Eliminating t between the following two equations, x-x_0=(v_0\cos\theta_0)t and y-y_0=(v_0\sin\theta_0)t-\tfrac{1}{2}gt^2, we obtain the equation of the path (the trajectory) of the projectile:

y=x\tan(\theta)-\frac{x^2g}{2v^2_{0}cos^2 \theta}

Time to reach the maximum height

t={v_{0y}\over g}

Time to reach ground

t={2v_{0y}\over g}

Displacement in X direction

\Delta x={v_{0x}t}

Displacement in Y direction

\Delta y={v_{0y}t-{\tfrac{1}{2}}gt^2}

Range of projectile

{r=\frac{v^2}{g}\sin(2\theta)}\,

Maximum height

y_{max}=\over 2g}

## Parabolic trajectory

Since g, \theta_0, and v_0 are constants, the above equation is of the form

y=ax+bx^2,

in which a and b are constants. This is the equation of a parabola, so the path is parabolic.

The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial height:

Note that R has its maximum value when \sin 2\theta_0=1, which does not necissarily corresponds to 2\theta_0=90^\circ or \theta_0=45^\circ.

Motion control

Motion control is a sub-field of automation, in which the position and/or velocity of machines are controlled using some type of device such as a hydraulic pump, linear actuator, or an electric motor, generally a servo. Motion control is an important part of robotics and CNCmachine tools, however it is more complex than in the use of specialized machines, where the kinematics are usually simpler. The latter is often called General Motion Control (GMC). Motion control is widely used in the packaging, printing, textile, semiconductor production, and assembly industries.

## Overview

The basic architecture of a motion control system contains:

• A motion controller to generate set points (the desired output or motion profile) and close a position and/or velocity feedback loop.
• A drive or amplifier to transform the control signal from the motion controller into a higher power electrical current or voltage that is presented to the actuator. Newer "intelligent" drives can close the position and velocity loops internally, resulting in much more accurate control.
• An actuator such as a hydraulic pump, air cylinder, linear actuator, or electric motor for output motion.
• One or more feedback sensors such as optical encoders, resolvers or Hall effect devices to return the position and/or velocity of the actuator to the motion controller in order to close the position and/or velocity control loops.
• Mechanical components to transform the motion of the actuator into the desired motion, including: gears, shafting, ball screw, belts, linkages, and linear and rotational bearings.

The interface between the motion controller and drives it controls is very critical when coordinated motion is required, as it must provide tight synchronization. Historically the only open interface was an analog signal, until open interfaces were developed that satisfied the requirements of coordinated motion control, the first being SERCOS in 1991 which is now enhanced to SERCOS III. Later interfaces capable of motion control include Profinet IRT, Ethernet Powerlink, and EtherCAT.

Common control functions include:

• Velocity control.
• Position (point-to-point) control: There are several methods for computing a motion trajectory. These are often based on the velocity profiles of a move such as a triangular profile, trapezoidal profile, or an S-curve profile.
• Pressure or Force control.
• Trans-mutational vector mapping.
• Electronic gearing (or cam profiling): The position of a slave axis is mathematically linked to the position of a master axis. A good example of this would be in a system where two rotating drums turn at a given ratio to each other. A more advanced case of electronic gearing is electronic camming. With electronic camming, a slave axis follows a profile that is a function of the master position. This profile need not be salted, but it must be an animated function.

Question:Car A, traveling at 50 miles per hour and located 700 ft ahead of car B, decelerates at a constant rate of 5 ft/s^2. Car B has a speed of 40 miles per hour in the same direction as car A and is accelerating at 8 ft/s^2. How far does car B travel in order to pass car A? According to my professor's notes, the answer is supposedly 1419 ft. However, my textbook says otherwise, giving me an answer of 1220 ft. I've tried this question about a million times over the past week and the final answer I've managed to get was 1214.1 ft. Now I ask you, WHO IS RIGHT!!! I am so confused right now =(

Answers:use an equation of motion relating velocity, acceleration, and distance: v =v +2a[x-x ] v = Initial velocity x = initial displacement The trick here is to know that when B reaches A, the final velocities (or velocities at that moment) are equal. With this said, set two of the equations equal to eachother. B side = A side v +2a[x-x ] = v +2a[x-x ] I run into a problem with units... PM me if you wish for me to continue the work, but it should pan itself out.

Question:I've been trying to solve this question for DAYS, however I still haven't been able to solve it. Car A, traveling at 50 mph and located 700 ft ahead of car B, decelerates at a constant rate of 5 ft/s^2. Car B has a speed of 40 mph in the same direction as car A and is accelerating at 8 ft/s^2. How far does car B travel in order to pass car A? I really need some help with this question. It's driving me crazy!

Answers:You know initial velocity (V0), constant acceleration (A), and beginning position (X0) for both cars. Use the position equation for constant acceleration. X=(1/2)At^2+V0*t+X0 . . . Xa=(1/2)(-5)t^2+50*t+700 Xb=(1/2)(8)t^2+40*t Find t when Xa=Xb. Set the two equations equal to each other. -2.5t^2+50t+700=4t^2+40t 6.5t^2-10t-700=0 Use the quadratic formula t=11.17, -9.63 Only the positive time makes sense in this problem. At time t=11.17, Car B has traveled, Xb=4t^2+40t=4(11.17)^2+40(11.17)=945.9 ft