Equations of Motion - Research Proposal Example

Equations of Motion Student Name: Student No: Course: Faculty: University: Lecturer: Date of Submission: Table of Contents Introduction 2 Objectives 4 Theory 4 Experiment 5 Results 10 The graph for Displacement against Time 10 The Graph for Velocity against Time 12 The Graph of Acceleration against Time 13 Conclusion 17 Works Cited 17 Introduction Various components of a moving object can be described by use of equations of motion. The kinematic principles such as velocity, displacement, acceleration and time can also be derived from the same equations. From the fundamental laws of constant acceleration, there equations are normally referred. These laws are applied in situations where acceleration is constant, and the motion of the object is in a straight line. The three equations of motion are: (a) (b) ; and, (c) s = Where; a = acceleration (m/s²) t = time (s) u = initial velocity (m/s) v = final velocity (m/s) s = displacement (m) From the equations, the first one is based on velocity-time relationship, In other words, when acceleration is constant, the rate of change of velocity is constant. If acceleration occurs for long, then the change of velocity will be greater (William, 2009). Also, when acceleration is constant, the rate at which velocity is changing becomes directly proportional to time but if the object is not accelerating, the initial and final velocities will be equal. In the second equation, it is based on the relationship between displacement and velocity. If the initial velocity is equal to zero, then the acceleration gets constant, and as a result, displacement is equated to square of velocity. In the third equation, the relationship of time and displacement is highlighted. A change in object displacement is directly proportional to time squared if the initial velocity is zero and acceleration is constant. The same equations of motion also apply to a projectile. In case of projectile, there are two forces that act on an object, which are the force of air resistance and gravity. Gravitational force influences vertical velocity. It has a constant value of 9.81 m/s2. Further, horizontal velocity of a projectile is influenced by air resistance (Thomas & Sherwood, 1972). At times it is ignored because its value is negligible. If a projectile object is released and then lands at same height, then it implies that both its initial and final velocities are equal. Also, the time taken to reach the top or apex is equated to half of the total time for the motion. That is, Peak height = ½ t if both the initial and final velocities are same. In finding the peak height, we take final velocity to be zero (Niels & Mads, 1996). In other words, a projectile changes direction when velocity becomes zero. Also, if we drop an object from a given height, we take the initial velocity to be zero. Objectives The objectives of this report are: To validate and demonstrate the applications of the equations of motion To use the equations of motion in calculating time, acceleration, velocity, and displacement Theory The equations of motions are simple mathematical formulas used to describe acceleration or velocity of an body in relation to some frame of reference. In classical mechanics, the fundamental equation of motion is F = ma (Lindberg & Florence, 1987). This is obtained from the second law of Newton where force acting on a body is equated to mass multiplied by acceleration (American Institute of Physics, 2005). If we know the force that is acting on a body as a function of time, then both the position and velocity of the same body can theoretically be derived as a function of time. This is achieved through the process of integration. For instance, if a falling object or body is accelerating at constant acceleration due to gravity, then velocity can be obtained by way of integrating the equation for acceleration whereas displacement is also obtained by integrating velocity equation (Dhont, 1996). In short, double integration of acceleration will give us the theoretical value of displacement. The reverse of this process is also true. Velocity and acceleration can be obtained theoretically by use of the process of differentiation. Experiment If we consider a car moving on a straight line at a constant acceleration, the equations of motion can be derived as shown below: a) The first equation of motion is given by: In this case, we can consider a body that is initially moving at velocity "Vi". The velocity of this body becomes "Vf" after some time interval "t". Therefore, the velocity change is given by: Vf - Vi or Due to changes in velocity, then acceleration “a” is also produced in the same body. The acceleration is given by: a = change in velocity /t If we replace the value of "V" then: , thus Or b) The second equation of motion is given by: S = Vit + 1/2at2 In this case, we consider a car that is moving on a road that is straight, and the initial or start velocity is ‘Vi’. The velocity, after time interval “t” becomes “Vf’. Therefore, the average velocity of the body is determined by: Average velocity = (Initial velocity + final velocity)/2 Or But, Vf = Vi + at If we replace the value of “Vf” in the equation: Average velocity (Vav) = (Vi + Vi + at)/2 = (2Vi + at)/2     = 2Vi/2 + at/2    Vav = Vi + at/2       Thus, Vav = Vi + 1/2at……………………………………….(i) However, the value of displacement, S, is given by: S = Vav x t By putting it in equation (i) above we find: S=[Vi +1/2at] t c) The third equation of motion is given by: 2aS = Vf2 – Vi2 Final velocity, initial velocity, distance and acceleration are related in 3rd equation of motion. Let’s consider a body that is moving at a velocity ‘Vi’. This velocity becomes ‘Vf’ after some interval of time (t). As a result of the changes in velocity, the body produces an acceleration ‘a’, assuming the distance travelled is ‘s’ meters. From the 1st equation of motion: Final velocity, Vf = Vi + at      Thus, (Vf – Vi)/a = t....................(ii) Average body velocity is given by: Average velocity, Vav = (Initial velocity + Final velocity)/2                            = (Vi + Vf)/2.................. (ii) But, displacement, S, is given by: S = Vav x t.................. (iv) Inserting the values of Vav & time (t) from equation (i) & (ii) in equation (iv) S = [ (Vf + Vi)/2] [ (Vf – Vi)/a], Thus, 2aS = (Vf + Vi)(Vf – Vi) Using the law, a2-b2 = (a+b)(a-b) These equations of motion can be used for drawing the graphs for finding velocity, acceleration and displacement if acceleration is a constant. The graph below shows the relationship between velocity, time, displacement and acceleration. The symbols for displacement, initial velocity, etc. are shown on the diagram. Thus, average speed is given by: Average speed = The distance travelled, s, is also given by : average speed x t therefore, Thus, at = v - u and v = u + at Results The graph for Displacement against Time For this graph, we consider a car that is travelling at constant velocity on a straight road. The total displacement of the car is measured every second, and the results are tabulated in the table below: t/s s/m 0 0 1 5 2 10 3 15 4 20 If the corresponding date points for this table are draw, a plot show below is obtained. By analysis the graph above, the slope is given by: m = change in s = 20 m = 5 m.s-1 Change in t 4 s The value of slope is velocity. In other words, velocity represents slope because it is obtained by dividing displacement (s) and time (t). The graph equation is given by: s = v x t, in this equation, v is simply a constant of proportionality Finding the area under the graph: Area = ½ × 4 s × 20 m = 40 m.s. Since the units do not stand for any physical quantity, the area under this kind of graph does not stand for any physical quantity. Thus, the gradient (slope) of displacement time graph represents velocity. The Graph for Velocity against Time For this plot, we consider a car that starts from rest, & then gains speed at a uniform manner (constant acceleration) along a straight line or road. The table below shows the values obtained for instantaneous velocity measured every second. t/s v/m.s-1 0 0 1 2 2 4 3 6 4 8 Graph of velocity against time The slope of this graph is given by: m = change in v = (8 m.s-1 )/4= 2 m.s-2 change in t The units of the slope are those of acceleration. Therefore, it is obvious that slope stands for acceleration because the slope is obtained by dividing velocity by time. The graph equation is v = a x t where is a constant of proportionality. The area under the graph is given by: ½ × 4 s × 8 m.s-1 = 16 m. Thus, the units are in meters and this represents displacement. The area under the graph therefore represents displacement. The Graph of Acceleration against Time The study of equations of motions is limited to only uniform acceleration (Clifford, 1993). The slopes for all the graphs is zero, it shows that accelerations is changing at a zero rate. The area under the graph should be evaluated. THE R Relationship Between Acceleration, Velocity And Displacement If a car is undergoing a constant acceleration of, say, 4 m.s-2 from a point of rest, the following table can be used to illustrate the relationship between acceleration, velocity, and time. Total time elapsed (t) in sec t 0 1 2 3 4 5 Final velocity, v in m.s-1 v = u + at 0 4 8 12 16 20 Average velocity, VAv at end of interval / m.s-1 vav = 1/2(u + v) 0 2 6 10 14 18 Displacement per interval m s = ½(u + v)t 0 2 6 10 14 18 Total displacement (m) s = ut + ½at2 0 2 8 18 32 50 Worked Examples to Illustrate the Application of the Equations of Motion Example 1: If we consider a car that is travelling on a straight road and has an initial velocity of 2 m/s, accelerating at 5 seconds and final velocity equal to 7 m/s, then: The acceleration during the first seconds Average velocity of the car during the 5 seconds Displacement of the car in 5 seconds (a) a = (v – u)/2 =( 7 m.s-1 - 2 m.s-1)/5 = 1 m.s-2 in the direction of motion. (b) vav = ( u + v )/t = (2 m.s-1 + 7 m.s-1)/2 = 4,5 m.s-1 in the direction of motion. (c) s = vav.t = (4,5 m.s-1)(5 s) = 22.5 m.s-1 in the direction of motion. Alternatively, = 22.5 m in the direction of motion Or s = (v2 - u2 )/2a={ (7 m.s-1)2 - (2 m.s-1)2 }/ (2 x1 )= = 22.5 m in the direction of motion Example 2: In this example, we consider a car accelerating at a uniform speed from a point of rest, attains maximum velocity of 30 m.s-1 in 10 seconds, and then travels at the same velocity more 150m. The brakes are applied and the car is made to decelerate at a uniform acceleration of -5 m.s-2 till it rests. The following workings illustrate how equations of motions are applied in calculation of: The car’s initial acceleration The time the car takes before coming to rest from the time the brakes are applied The overall time taken when the car is in motion And the total displacement covered by the car (a) a1 =( v1 - u1)/ t1 = {(30 - 0)m.s-1 }/10s= 3 m.s-2 in the direction of motion. (b) t3 =( v3 - u3) a3 ={ (0 - 30)m.s-1 }/5= -6 s (c) Time covered from 150 m at 30 m.s-1: t2 = s2 =   150 m/30m/s  = 5 s tTotal = (d) s1 = ½a1t12 = ½(3 m.s-2)(10 s)2 = 150 m s3 = u3t3 + ½a3t32 = (30 m.s-1)(6 s) + ½(-5 m.s-2)(6 s)2 = 90 m sTotal = s1 + s2 + s3 = 150 m + 150 m + 90 m = 390 m in the direction of motion. Conclusion The equations of motions can be applied in solving calculations to do with different kinematic variables in cases where the acceleration is constant, and also when an object is travelling a line that is straight. If an object is project, then the gravitational force becomes a constant of acceleration. Works Cited American Institute of Physics (2005). Journal of Experimental and Theoretical Physics. Wiley: New York. Clifford, M.W. (1993). Theory and Experiment in Gravitational Physics. Press Syndicate of Cambridge University: Cambridge. Dhont, J.K. (1996). An Introduction to Dynamics of Colloids. Elsevier Science: Amsterdam. Lindberg, H.E., & Florence, A.L. (1987). Dynamic Pulse Buckling: Theory and Experiment. Martinus Nijhoff Publishers: Dordrecht. Niels, LP, & Mads, L. P. (1996). A Direct Derivation of the Equations of Motion for 3-D Flexible Mechanical Systems . Danish Center for Applied Mathematics and Mechanics: Denmark. Thomas, G.G., & Sherwood, H. (1972). Summary of Transformation Equations and Equations of Motion Used in Free-Flight and Wind-Tunnel Data Reduction and Analysis. SAGE Publications, New Jersey. William, M. (2009). Equations of Motion: Adventure, Risk and Innovation. Bentley Robert Incorporated: New York. Read More