Mathematics
Variable Acceleration
Variable acceleration refers to a situation where an object's rate of change of velocity is not constant. In mathematical terms, this is represented by a function of time for acceleration. This concept is commonly encountered in physics and engineering, where the motion of objects is often influenced by varying forces or conditions.
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7 Key excerpts on "Variable Acceleration"
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
Average velocity is a vector that has the same direction as the displacement. When the elapsed time becomes infinitesimally small, the average velocity becomes equal to the instantaneous velocity v B , the velocity at an instant of time, as indicated in Equation 2.3. 2.3 Acceleration The average acceleration a B is a vector. It equals the change D v B in the velocity divided by the elapsed time Dt, the change in the velocity being the final minus the initial velocity; see Equation 2.4. When Dt becomes infinitesimally small, the average acceleration becomes equal to the instantaneous acceleration a B , as indicated in Equation 2.5. Acceleration is the rate at which the velocity is changing. 2.4 Equations of Kinematics for Constant Acceleration/2.5 Applications of the Equations of Kine- matics The equations of kinematics apply when an object moves with a constant acceleration along a straight line. These equations relate the displacement x 2 x 0 , the acceleration a, the final velocity v, the initial velocity v 0 , and the elapsed time t 2 t 0 . Assuming that x 0 5 0 m at t 0 5 0 s, the equations of kinematics are as shown in Equations 2.4 and 2.7–2.9. 2.6 Freely Falling Bodies In free-fall motion, an object experiences negligible air resistance and a constant acceleration due to gravity. All objects at the same location above the earth have the same acceleration due to gravity. The acceleration due to gravity is directed toward the center of the earth and has a magnitude of approximately 9.80 m/s 2 near the earth’s surface. 2.7 Graphical Analysis of Velocity and Acceleration The slope of a plot of position versus time for a moving object gives the object’s velocity. The slope of a plot of velocity versus time gives the object’s acceleration.
eBook - PDFWorkshop Physics Activity Guide Module 1
Mechanics I
- Priscilla W. Laws, David P. Jackson, Brett J. Pearson(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
We will explore the motion of objects near the surface of Earth much more thoroughly in a later unit. For now we simply note that we have seen two situations in which the acceleration is observed to be constant. Although most everyday motions do not exhibit constant acceleration, it is a common enough occurrence to devote some time to examining this specific type of motion. 4.4 EQUATIONS DESCRIBING CONSTANT ACCELERATION As we have seen, quantitative information about velocity and acceleration is contained in a position-time graph. Similarly, the equations describing position, velocity, and acceleration are also related. In this section we derive the equations that describe the motion of an object moving with constant acceleration (includ- ing the possibility of zero acceleration). Velocity for Constant Acceleration We previously discussed the concept of average acceleration, or the change in velocity over a specified time interval. For one-dimensional motion, we need only concern ourselves with the x-components of the vectors (assuming the motion is in the x-direction), so the average and instantaneous accelerations are given by: ⟨a x ⟩ = Δ x Δt = 2x − 1x t 2 − t 1 and a x = lim t→0 Δ x Δt = d x dt (4.1) where the subscripts 1 and 2 indicate arbitrary initial and final points, respec- tively. Normally, the instantaneous acceleration is different from the average acceleration; however, a constant acceleration always has the same value, so there is no distinction between the two: ⟨a x ⟩ = a x . Thus, for constant acceler- ation we can refer to a x simply as the acceleration. 4.4.1. Activity: Deriving the Velocity Equation a. Let’s assume we start monitoring the motion of an object at some initial time t 0 . (In physics it is common to use t 0 , pronounced “t-zero” or “t-naught,” as the initial time.) The initial velocity in the x-direction at this time is then labeled as 0x (“v-zero” or “v-naught”).
eBook - PDFGeneral Engineering Science in SI Units
In Two Volumes
- G. W. Marr, N. Hiller(Authors)
- 2016(Publication Date)
- Pergamon(Publisher)
The definition of acceleration indicates that a unit of acceler-ation is obtained from the ratio u ^uoft^me y · T h u s i n t h e SI s y s t e m the basic unit is the metre s ^ n s d econd or metre per second per second, usually abbreviated to m/s 2 . EXAMPLE. A cyclist, starting from rest, attains a speed of 18 km/h in 55 s, while travelling in a straight line. Calculate his aver-age acceleration. Change in velocity = 18 — 0 = 18 km/h 18X10 3 , = -T6ÖÖ-m/s = 5 m/s Time required for change = 55 s Λ average acceleration = ^ m/s 2 ^ 0091 m/s 2 EXAMPLE. A train pulls out of a station with a uniform accelera-tion of 15 cm/s 2 . How long will it take to reach a speed of 54 km/h? Change in speed = 54 — 0 km/h = 54 km/h 54X10 3 3 600 = 15 m/s m/s A Λ . change in speed Acceleration = %— Λ change in speed • · * im e required = acceleration 15 m/s 15XlO-2 m/s 2 100 s 36 VELOCITY AND ACCELERATION 2.4. Distance-time Graphs A distance-time graph is a graph in which the distance travelled in a given time is plotted as ordinate, the corresponding time being plotted as abscissa. For non-uniform motion the graph will be a curve, such as that shown in Fig. 2.2. A general idea of the motion can be deduced from such a graph. Over the portion OA of the graph the curve is concave upward; that is, the graph is W QJ tt> E. w m to u c n V) Q 80 70 bO 50 40 30 .20 10 u 10 20 30 40 50 Time, t seconds F I G . 2.2. rising more and more steeply. The distance travelled during any particular time interval within this total period is thus greater than the distance travelled during the preceding equal interval of time. Hence the average velocity is increasing and the motion is therefore accelerated motion. Over the portion AB, the concavity of the curve is downward; the velocity is progressively decreasing and therefore the moving body is decelerating. During the time interval (i 2 — h), the distance travelled is O2—$i), as indicated in Fig. 2.3. The average velocity during this period is 37
eBook - PDFGeneral Engineering Science in SI Units
The Commonwealth and International Library: Mechanical Engineering Division
- G. W. Marr, N. Hiller(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
The definition of acceleration indicates that a unit of acceler-ation is obtained from the ratio u ^ ^ ^ f . Thus in the SI system the basic unit is the metre J^^ econd or metre per second per second, usually abbreviated to m/s 2 . EXAMPLE. A cyclist, starting from rest, attains a speed of 18 km/h in 55 s, while travelling in a straight line. Calculate his aver-age acceleration. Change in velocity = 18 — 0 = 18 km/h 18X10 3 , = -3 W m / s = 5 m/s Time required for change = 55 s Λ average acceleration = / 5 m/s 2 Δ 0091 m/s 2 EXAMPLE. A train pulls out of a station with a uniform accelera-tion of 15 cm/s 2 . How long will it take to reach a speed of 54 km/h? Change in speed = 54 — 0 km/h = 54 km/h 54X10 3 . = -3 6ÖÖ~ m / s = 15 m/s A , Λ . change in speed Acceleration = 7 7 — time , · · , change in speed • · t i m e required = accelcration 15 m/s 15X10-2 m/s 2 100 s 36 VELOCITY AND ACCELERATION 2.4. Distance-time Graphs A distance-time graph is a graph in which the distance travelled in a given time is plotted as ordinate, the corresponding time being plotted as abscissa. For non-uniform motion the graph will be a curve, such as that shown in Fig. 2.2. A general idea of the motion can be deduced from such a graph. Over the portion OA of the graph the curve is concave upward; that is, the graph is 20 30 40 50 Time, t seconds FIG. 2.2. rising more and more steeply. The distance travelled during any particular time interval within this total period is thus greater than the distance travelled during the preceding equal interval of time. Hence the average velocity is increasing and the motion is therefore accelerated motion. Over the portion AB, the concavity of the curve is downward; the velocity is progressively decreasing and therefore the moving body is decelerating. During the time interval (/ 2 —Ί), the distance travelled is (^ 2 — ·?ι), as indicated in Fig. 2.3. The average velocity during this period is 37
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
When ∆t becomes infinitesimally small, the average acceleration becomes equal to the 48 CHAPTER 2 Kinematics in One Dimension instantaneous acceleration a → , as indicated in Equation 2.5. Acceleration is the rate at which the velocity is changing. a → = ∆v → ∆t (2.4) a → = lim ∆ t →0 ∆ v → ∆t (2.5) 2.4 Equations of Kinematics for Constant Acceleration/2.5 Applications of the Equations of Kinematics The equations of kinematics apply when an object moves with a constant acceleration along a straight line. These equa- tions relate the displacement x − x 0 , the acceleration a, the final velocity υ, the initial velocity υ 0 , and the elapsed time t − t 0 . Assuming that x 0 = 0 m at t 0 = 0 s, the equations of kinematics are as shown in Equations 2.4 and 2.7–2.9. υ = υ 0 + at (2.4) x = 1 2 (υ 0 + υ) t (2.7) x = υ 0 t + 1 2 at 2 (2.8) υ 2 = υ 2 0 + 2ax (2.9) 2.6 Freely Falling Bodies In free-fall motion, an object experiences neg- ligible air resistance and a constant acceleration due to gravity. All objects at the same location above the earth have the same acceleration due to gravity. The acceleration due to gravity is directed toward the center of the earth and has a magnitude of approximately 9.80 m/s 2 near the earth’s surface. 2.7 Graphical Analysis of Velocity and Acceleration The slope of a plot of position versus time for a moving object gives the object’s velocity. The slope of a plot of velocity versus time gives the object’s acceleration. Note to Instructors: The numbering of the questions shown here reflects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS. Section 2.1 Displacement 1. What is the difference between distance and displacement? (a) Distance is a vector, while displacement is not a vector. (b) Displacement is a vector, while distance is not a vector.
eBook - PDFTheoretical Mechanics for Sixth Forms
in Two Volumes
- C. Plumpton, W. A. Tomkys(Authors)
- 2017(Publication Date)
- Pergamon(Publisher)
CHAPTER XVII MOTION WITH Variable Acceleration 17.1. Kinematics In Chapter III of Volume I we discussed the graphical relationships among the quantities distance, velocity, acceleration and time for a par-ticle moving in a straight line. Graphical interpretations of the defini-tions ds _ du _ du n_ dt ' f dt - n ds (in the usual notation) enabled us to estimate any two of the quantities s, v, f and t within the range involved if a number of corresponding values of the other two were known. In each of the examples which follow a functional relationship between two of the quantities is known, necessary boundary conditions are given, and the remaining quantities are calculated. Example 1. A particle moves in a straight line so that at time t from the beginning of the motion its displacement from a fixed point O in the line is s and its velocity is v where v = ks, k being constant, and s = a when t = O. Find (a) the initial acceleration of the particle, (b) the value of s when t = 2 seconds, (c) the equation of motion of the particle relating v and t. (a) From v = ks, dv/ds = k and the acceleration of the particle is n ds = k 2 s. Therefore, since s = a when t = 0, the initial acceleration of the particle is k 2 a. 487 488 THEORETICAL MECHANICS ds n dt = ks. J s J lns =kt +A, where A is constant. But s=a when t =0; A=In a. In s—In a= kt. In~sl = kt. a j s = a e u , Therefore, when t = 2 seconds, s = a e lk . (c) First Method. v = ks and s = a e k '. Eliminating s we have v = ka ex: Second Method. s = a e k '. v= ds = ka e'. Example 2. In the usual notation, the equation of motion of a particle moving in a straight line is f = k/s 2 , for s z 1, and s = 1,1 = 0 when t = O. Show that v cannot exceed ß/(2k) and find the time taken by the particle to move from s = 2 to s = 4. dv k n = ds s 2 ,' . J v dv = J k ds . s n ~ k . 2 A s , where A is constant. But s =lwhenv =0; A=k. • n2 = 2k (1-1 1, s / i.e. v = {2k (1 — s ){ and therefore v - ß/(2k) for all values of s.
eBook - PDF- Jerry B. Marion(Author)
- 2013(Publication Date)
- Academic Press(Publisher)
In order to do this, we require vectors to represent the position, the velocity, and the acceleration of a given particle. It is customary to specify the position of a particle with respect to a certain reference frame by a vector r, which in general will be a function of time : r = r(i). The velocity vector v and the acceleration vector a are defined according to 2.3 EXAMPLES OF DERIVATIVES—VELOCITY AND ACCELERATION 35 where a single dot above a symbol denotes the first time derivative and two dots denote the second time derivative. In rectangular coordinates the expressions for r, v, and a are r -x l e 1 + x 2 e 2 + x 3 e 3 = Σ χ & ( 2 · 7 ) i v = r = Xx i e i = X ^ e i (2.8) a = v = r = £ * A = Z § e < (2.9) The calculation of these quantities in rectangular coordinates is straight-forward since the unit vectors e f are constant in time. In nonrectangular coordinate systems, however, the unit vectors at the position of the particle as it moves in space are not necessarily constant in time, and the compo-nents of the time derivatives of r are no longer simple relations as above. We shall not have occasion to discuss general curvilinear coordinate systems here, but plane polar coordinates, spherical coordinates, and cylindrical coordinates are of sufficient importance to warrant a discussion here of velocity and acceleration in these coordinate systems.* In order to express v and a in plane polar coordinates, consider the situation in Fig. 2-2. Here, a point moves along the curve r(i) and in the time interval t 2 — t 1 = dt moves from P {1) to P {2 The unit vectors, e r and e 0 , which are orthogonal, change from ë r 1] to e< 2) and from e e l) to e ( e 2 The change in e r is e< 2 > -β < υ = de r (2.10) which is a vector normal to e r (and, therefore, in the direction of e 0 ). Simi-larly, the change in e e is # > -# > = de e (2.11) which is a vector normal to e e .
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