Constant Acceleration
Constant acceleration refers to a situation where an object's velocity changes by the same amount in each equal time interval. In mathematical terms, it means that the rate of change of velocity with respect to time remains constant. This concept is often used in physics and engineering to analyze the motion of objects under the influence of a constant force.
8 Key excerpts on "Constant Acceleration"
eBook - ePub- Alan Hendrickson(Author)
- 2012(Publication Date)
- Routledge(Publisher)
...2 The Equations of Constant Acceleration DOI: 10.4324/9780080557540-2 Constant Acceleration Acceleration is a measure of how rapidly an object speeds up or slows down, or more specifically, how much its velocity changes over some given interval of time. Acceleration can be measured both at a point in time, the result being called instantaneous acceleration, or over some finite time interval, called an average acceleration. Constant Acceleration exists whenever both measures of acceleration remain the same for some period of time. During these time periods, average acceleration, ā, and instantaneous acceleration, a, are equal, and both hold at one value regardless of the length of the time interval chosen for Δ t : a = a ¯ = Δ v Δ t i f a c c e l e r a t i o n i s c o n s t a n t In a typical scenery move, there are three key periods where acceleration can be considered constant: during acceleration from zero on up to top speed, during travel at a constant top speed, and during deceleration to a stop (see motion profiles in Figure 2.1). During the constant velocity portion, acceleration, a, will be zero because velocity does not change. (True too before and after the move, but since nothing is moving then, it is of no interest to us here.) During a Constant Acceleration or deceleration, the change of velocity versus time is a fixed number regardless of the value chosen for Δ t. For three arbitrary values of t 1 and t 2 in the example in Figure 2.1, acceleration calculates out to the same value: Figure 2.1 Typical Constant Acceleration motion. profiles f o r t 1 = 1 a n d t 2 = 2 a ¯ = v 2 − v 1 t 2 − t 1 = 1 − 0 2 − 1 = 1 f t / s e c 2 f o r t 1 = 1 a n d t 2 = 5 a ¯ = v 2 − v 1 t 2 − t 1 = 4 − 0 5 − 1 = 1 f t /[--=PL...
eBook - ePub- Richard Gentle, Peter Edwards, William Bolton(Authors)
- 2001(Publication Date)
- Newnes(Publisher)
...We will see in the next section that this means that the driver will have to exert a force on the steering wheel which could be calculated. Often we take speed and velocity as the same thing and use the symbols v or u for both, but do not forget that there is a distinction. If the object is moving at a constant velocity v and it has travelled a distance of s in time t then the velocity is given by (4.1.1) Alternatively, if we know that the object has been travelling at a constant velocity v for a time of t then we can calculate the distance travelled as (4.1.2) Acceleration is the rate at which the velocity is changing with time and so it is defined as the change in velocity in a short time, divided by the short time itself. Therefore the units are metres per second (the units of velocity) divided by seconds and these are written as metres per second 2 (m/s 2). Acceleration is generally given the symbol a. Usually the term acceleration is used for the rate at which an object’s speed is increasing, while deceleration is used when the speed is decreasing. Again, do not forget that a change in velocity could also be a directional change at a constant speed. Having defined some of the common quantities met in the study of kinematics, we can now look at the way that these quantities are linked mathematically. Velocity is the rate at which an object’s displacement is changing with time. Therefore if we were to plot a graph of the object’s displacement s against time t then the value of the slope of the line at any point would be the magnitude of the velocity (i.e. the speed). In Figure 4.1.1, an object is starting from the origin of the graph where its displacement is zero at time zero. The line of the graph is straight here, meaning that the displacement increases at a constant rate. In other words, the speed is constant to begin with and we could measure it by working out the slope of the straight line portion of the graph...
eBook - ePub- W. Bolton(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...If the velocity changes from 5 m/s to 2 m/s in 10 s then the average acceleration over that time is (22 5)/10 5 20.3 m/s 2, i.e. it is a retardation. 11 A constant or uniform acceleration occurs when the velocity changes by equal amounts in equal intervals of time, however small the time interval. Thus an object with a Constant Acceleration of 5 m/s 2 in a particular direction for a time of 30 s will change its velocity by 5 m/s in the specified direction in each second of its motion. 4.2 Straight line motion The equations that are derived in the following discussion all relate to uniformly accelerated motion in a straight line. If u is the initial velocity, i.e. at time t = 0, and v the velocity after some time t, then the change in velocity in the time interval t is (v - u). Hence the acceleration a is (v 2 u)/ t. Rearranging this gives: v = u + at [Equation 1] If the object, in its straight- line motion, covers a distance s in a time t, then the average velocity in that time interval is s/t. With an initial velocity of u and a final velocity of v at the end of the time interval, the average velocity is (u + v)/2. Hence s/t 5 (u 1 v)/2 and so: Substituting for v by using the equation v = u + at gives: Consider the equation v = u + at. Squaring both sides of this equation gives: Hence, substituting for the bracketed term using equation 2: v 2 = u 2 + 2as [Equation 3] The equations [1], [2] and [3] are referred to as the equations for straight- line motion. The following examples illustrate their use in solving engineering problems. Example An object moves in a straight line with a uniform acceleration. If it starts from rest and takes 12 s to cover 100 m, what is the acceleration? If it continues with the same acceleration, how long will it take to cover the next 100 m and what will be its velocity after the 200 m? For the first 100 m, we have u = 0, s = 100 m, t = 12 s and are required to obtain a...
eBook - ePub- John Bird(Author)
- 2019(Publication Date)
- Routledge(Publisher)
...Chapter 18 Acceleration Why it is important to understand: Acceleration Acceleration may be defined as a ‘change in velocity’. This change can be in the magnitude (speed) of the velocity or the direction of the velocity. In daily life we use acceleration as a term for the speeding up of objects and decelerating for the slowing down of objects. If there is a change in the velocity, whether it is slowing down or speeding up, or changing its direction, we say that the object is accelerating. If an object is moving at constant speed in a circular motion – such as a satellite orbiting the earth – it is said to be accelerating because change in direction of motion means its velocity is changing even if speed may be constant. This is called centripetal (directed towards the centre) acceleration. On the other hand, if the direction of motion of the object is not changing but its speed is, this is called tangential acceleration. If the direction of acceleration is in the same direction as that of velocity then the object is said to be speeding up or accelerating. If the acceleration and velocity are in opposite directions then the object is said to be slowing down or decelerating. An example of Constant Acceleration is the effect of the gravity of earth on an object in free fall. Measurement of the acceleration of a vehicle enables an evaluation of the overall vehicle performance and response. Detection of rapid negative acceleration of a vehicle is used to detect vehicle collision and deploy airbags. The measurement of acceleration is also used to measure seismic activity, inclination and machine vibration...
eBook - ePub- A. D. Johnson(Author)
- 2017(Publication Date)
- Routledge(Publisher)
...The motion can be expressed graphically as shown in Figure 2.5. Fig. 2.5 Uniform acceleration shown on a velocity–time graph. Referring to Figure 2.5, the following observations can be made: change of velocity = v 2 − v 1 This change of velocity can be related to the acceleration because acceleration = a = rate of change of velocity = changeofvelocity timetaken = v 2 − v 1 t and transposing gives v 2 = v 1 + a t (2.2) This equation can be compared directly with the diagram in Figure 2.5. Figure 2.5 shows that the velocity increases at a uniform rate between v 1 and v 2. The average velocity can therefore be described as v ¯ = (v 1 + v 2) 2 and if s is the distance travelled during that period, equation (2.1) may be used in the form s = v t = average velocity × time In other words s = v ¯ × t and s = (v 1 + v 2) t 2 (2.3) From equation (2.2), it is possible to substitute for υ 2 in equation (2.3) : s = (v 1 + v 1 + a t) t 2 and rearranging. gives s = v 1 t + 1 2 a t 2 (2.4) The above equations are adequate for most purposes but there often arise situations where time t is the unknown. Since equations (2.2), 2.3 and (2.4) all possess t an expression is now needed for v 2 in terms of v 1, a and s, which excludes t. This is done by squaring both sides of equation (2.2) and substituting for s from equation (2.4). From equation (2.2) v 2 = v 1 + a t and squaring gives v 2 2 = (v 1 + a t) 2 so that v 2 2 = v 1 2 + 2 v 1 a t + a 2 t 2 or v 2 2 = v 1 2 + 2 a (v 1 t + 1 2 a t 2) But (v 1 t + 1 2 a t 2) is equal to s from equation. (2.4) : v 2 2 = v 1 2 + 2 a s (2.5) The equations (2.2), 2.3, 2.4 and (2.5), derived above, form the basis of all motion studies at this level and can easily be manipulated to cope with a range of situations, including falling bodies and rotary motion, as explained later. Example 2.3 A train has a uniform acceleration of 0.2 m/s 2 along a straight track...
eBook - ePubInstant Notes in Sport and Exercise Biomechanics
Second Edition
- Paul Grimshaw, Michael Cole, Adrian Burden, Neil Fowler(Authors)
- 2019(Publication Date)
- Garland Science(Publisher)
...The situation is more complicated when a body moves in two directions, again in a straight line. An example of this is when someone jumps directly up and then lands back in the same place (e.g. an SVJ) and experiences the Constant Acceleration due to gravity during both the ascent and descent. In this situation the velocity of the body decreases linearly to zero at the apex of the jump and then increases in the same manner until landing. Their position changes in a curvilinear fashion (Figure A6.2). The changes in position and velocity of a constantly accelerating body were first noted by an Italian mathematician named Galileo in the early seventeenth century. Galileo also derived the following equations of uniformly accelerated motion that can be used to generate the curves shown in Figure A6.1 and Figure A6.2, and therefore to describe the motion of bodies experiencing Constant Acceleration. v 2 = v 1 + at (Equation A6.1) d = v 1 t + ½ at 2 (Equation A6.2) v 2 2 = v 1 2 + 2ad (Equation A6.3) d = ½ (v 1 + v 2)t (Equation A6.4) Figure A6.1 Horizontal position and velocity of a body when experiencing a Constant Acceleration Figure A6.2 Vertical position, velocity and acceleration of the centre of mass (COM) during the flight phase of a standing vertical jump where: v 1 = initial velocity v 2 = final velocity d = change in. position or displacement t = change in time Sport and exercise biomechanists often wish to analyse the motion of a body while it experiences Constant Acceleration. It may be important to know, for example, how high somebody jumped, what velocity they would experience after a certain time, or how long it would take them to reach that velocity. Any of the equations above (A6.1–A6.4) can be used to answer such questions. For example, consider someone performing a SVJ with a take-off velocity of their centre of mass (COM) of 2.4 m/s. What would be the displacement of their COM between the instant of take-off and the highest point of their COM (i.e...
eBook - ePubAutomotive Accident Reconstruction
Practices and Principles, Second Edition
- Donald E. Struble, John D. Struble(Authors)
- 2020(Publication Date)
- CRC Press(Publisher)
...Therefore what-if questions may focus on the effects of varying the perception–reaction time from say a ½ to 2½ sec. Constant Acceleration The simplest kind of motion is at constant velocity, for which “rest” is simply a special case. Most often, though, time–distance studies involve vehicles that are being accelerated. The type of acceleration that is simplest to analyze is a constant, for which constant velocity (zero acceleration) is a special case. If a trajectory can be divided into segments with Constant Acceleration (as has been done in Chapters 3 and 4), the resulting analysis will cover the vast majority of questions asked of the reconstructionist. Generally, it is not necessary to make the segments small, but of course the velocity and the travel distance need to be continuous functions over an entire path. It is possible to write a general-purpose computer program in which different formulas are used in various segments, depending on what is known and what is unknown in any given segment. One can even do a forward-directed or a backward-directed calculation, depending on whether the velocity at the end is known (a final-value problem) or the velocity at the beginning is known (an initial-value problem). Obviously, a lot of IF-THEN-ELSE constructs must be employed. In all cases, the object is to wind up knowing the velocity at each of the segment boundaries, otherwise known as the key points. In addition, the user should have access to both the elapsed time (which is a forward measure) and the remaining time (which is a backward measure). Often the final event is a collision, though it could also be a vehicle coming to rest or even something else. Similarly, the user should know or be able to calculate the elapsed distance and the remaining distance. In each segment, the user should know or be able to calculate Δ S, the segment length, and Δ t, the segment time. The time–distance study can (and usually does) involve more than one vehicle...
eBook - ePub- Gregory Hill, Mel Friedman(Authors)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...It does not take into account direction of motion. Speed at any moment t = c is | v(c) | Acceleration If x (t) is the position of a particle moving along the x -axis, its acceleration The acceleration of the particle measured at any moment t = c, is a (c) = v ′(c) = x ″(c). EXAMPLE 5.19 The position of a particle moving along the x -axis is defined by x (t) = t · sin(2 t). Find the displacement of the particle on the time interval, t = 0 to seconds. SOLUTION The particle has oscillated during the interval, but has returned to its starting position when seconds, so there is no displacement. EXAMPLE 5.20 The position of a particle moving along the x -axis is defined by x (t) = l n (t 2 + 1) for t ≥ 0. If distance is measured in feet and time is in seconds, find the average acceleration on the time interval [1, 3] seconds. SOLUTION If average velocity is change in position over change in time, by extension, average acceleration is change in velocity over change in time. Velocity = v (t) Average acceleration EXAMPLE 5.21 The position of a particle moving along the x -axis is defined by For t ≥ 0, find when the particle is moving right. SOLUTION The particle is moving right when its velocity is positive. Find when the velocity is 0 and test the intervals between those times. 0< t <2 2< t <4 t > 4 v (1) = 6 v (3) = –6 v (5) = 30 The particle is moving right in the time intervals 0 < t < 2 and t > 4. EXAMPLE 5.22 A particle is moving along the x -axis. Figure 5.20 shows its velocity graph. On the interval [0, 6], when is the speed of the particle greatest? Figure 5.20 SOLUTION Since speed is the absolute value of velocity, speed is greatest at about t = 5 seconds, not 2 seconds. 5.5 OPTIMIZATION Optimization is one of the most interesting applications of derivatives. It applies maximizing and minimizing to practical problems, and it involves a wide diversity of problems...
