Motion with Constant Acceleration & Free Falling Bodies - PDF

Motion with constant acceleration, Freely Falling Bodies Motion With Constant Acceleration The simplest kind of accelerated motion is straight-line motion with constant acceleration. In this case the velocity changes at the same rate throughout the motion. eg: a falling body has a constant accele...

Motion with constant acceleration, Freely Falling Bodies Motion With Constant Acceleration The simplest kind of accelerated motion is straight-line motion with constant acceleration. In this case the velocity changes at the same rate throughout the motion. eg: a falling body has a constant acceleration A velocity-time (v-t) graph of straight-line motion with constant positive acceleration A velocity-time 𝒗𝒙 − 𝑡 graph of straight-line motion with constant positive x-acceleration 𝒂𝒙 The initial x-velocity 𝒗𝟎𝒙 is also positive in this case  The graph of x-velocity versus time, Since the x-acceleration is or 𝑣𝑥 − 𝑡 graph, has a constant slope constant, the graph 𝑎𝑥 − 𝑡 because the acceleration is constant, (graph of x-acceleration versus so this graph is a straight line time) is a horizontal line.  𝒗𝟐𝒙 −𝒗𝟏𝒙 the expression for average x-acceleration 𝑎𝑎𝑣−𝑥 is 𝒂𝒂𝒗−𝒙 = 𝒕𝟐 −𝒕𝟏 When the x-acceleration 𝑎𝑥 is constant, the average x-acceleration 𝑎𝑎𝑣−𝑥 for any time interval is the 𝒗𝟐𝒙 −𝒗𝟏𝒙 same as 𝑎𝑥. Then 𝒂𝒙 = 𝒕𝟐 −𝒕𝟏 Now we let 𝑡1 = 0 and let 𝑡2 be any later time t. We use the symbol 𝑣0𝑥 for the initial x-velocity at time 𝑡 = 0; the x-velocity at the later time t is 𝑣𝑥. 𝑣𝑥 −𝑣0𝑥 Then 𝑎𝑥 = 𝑡−0 𝑡ℎ𝑒 𝑡𝑜𝑎𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 𝑣𝑥 − 𝑣0𝑥 = 𝑎𝑥 𝑡  𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 The x-velocity 𝑣𝑥 at any time t then equals the initial x-velocity 𝑣0𝑥 (at 𝑡 = 0) plus the change in x- velocity 𝑎𝑥 𝑡  Next we’ll derive an equation for the position x as a function of time when the x-acceleration is constant which is true whether or not the acceleration is constant. We call the position at time 𝑡 = 0 the initial position, denoted by 𝑥0 and the position at 𝑡𝑖𝑚𝑒 𝑡 is simply 𝑥. Thus for the time interval ∆𝑡 = 𝑡 − 0 and the displacement is ∆𝑥 = 𝑥 − 𝑥0 , So 𝒙−𝒙𝟎 𝒙−𝒙𝟎 𝒗𝒂𝒗−𝒙 = = (2.9) 𝒕−𝟎 𝒕 If the x-acceleration is constant then that the x-velocity changes at a constant rate. In this case the average x-velocity (𝑣𝑎𝑣−𝑥 )for the time interval from 0 𝑡𝑜 𝑡 is simply the average of the x-velocities at the beginning (𝑣0𝑥 )and end of the interval (𝑣𝑥 ): 1 𝑣𝑎𝑣−𝑥 = (𝑣0𝑥 +𝑣𝑥 ) (for constant acceleration only) (2.10) 2 We also know that with constant x-acceleration, the x-velocity 𝑣𝑥 at any time 𝑡 is given by 𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 Substituting that expression for 𝑣𝑥 into Eq. (2.10), we find 1 𝑣𝑎𝑣−𝑥 = (𝑣0𝑥 +𝑣0𝑥 + 𝑎𝑥 𝑡) , (for constant acceleration only) 2 𝟏 𝒗𝒂𝒗−𝒙 = 𝒗𝟎𝒙 + 𝒂𝒙 𝒕 (2.11) 𝟐  𝒙−𝒙𝟎 𝒗𝒂𝒗−𝒙 = (2.9) 𝒕 𝟏 𝒗𝒂𝒗−𝒙 = 𝒗𝟎𝒙 + 𝒂𝒙 𝒕 (2.11) 𝟐 𝟏 𝒙−𝒙𝟎 𝒗𝟎𝒙 + 𝒂𝒙 𝒕 = 𝟐 𝒕 𝟏 𝒙 − 𝒙𝟎 = 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝒕𝟐 𝟐 𝟏 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝒕𝟐 (2.12) 𝟐 Equation (2.12) tells us that the particle’s position at time t is the sum of three terms: its initial position at t = 0, 𝒙𝟎 , plus the displacement 𝒗𝟎𝒙 𝒕 it would have if its x-velocity remained equal to its initial value 𝒗𝟎𝒙 , 𝟏 plus an additional displacement 𝒂𝒙 𝒕𝟐 caused by the change in x-velocity 𝟐  𝟏 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝟐 𝒂𝒙 𝒕𝟐 (2.12) A graph of Eq. (2.12)—that is, an 𝑥 − 𝑡 𝑔𝑟𝑎𝑝ℎ for motion with constant x-acceleration (Fig. 2.18a)—is always a parabola. Figure 2.18b shows such a graph. If there is zero x-acceleration, the x-t graph is a straight line; if there is a constant x-acceleration, the additional 𝟏 𝟐 𝟏 𝟐 𝒂 𝒙 𝒕 term in Eq. 𝒙 = 𝒙 𝟎 + 𝒗 𝟎𝒙 𝒕 + 𝒂𝒙 𝒕 𝟐 𝟐 for x as a function of t curves the graph into a parabola Similarly, if there is zero x-acceleration, the 𝒗𝒙 − 𝒕 graph is a horizontal line (the x-velocity is constant). Adding a constant x- acceleration in Eq. (2.8) gives a slope to the graph (Fig. 2.19b)  the change in x-velocity of the particle equals the area under the 𝑎𝑥 − 𝑡 graph, then the displacement (change in position) equals the area under the 𝑣𝑥 − 𝑡 graph. So the displacement 𝑥 − 𝑥0 of the particle between t = 0 and any later time t equals the area under the 𝑣𝑥 − 𝑡 graph between those times.  we divide the area under the graph into a dark-colored rectangle (vertical side 𝑣0𝑥 , horizontal side 𝑡, and area 𝑣0𝑥 t) and a light- colored right triangle (vertical side 𝑎𝑥 𝑡 horizontal side 𝑡, and area 1 1 2 𝑡 𝑎 𝑥 𝑡 = 𝑎 𝑥 𝑡 ). 2 2 1 The total area under the 𝑣𝑥 − 𝑡 graph is 𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 + 2 𝑎𝑥 𝑡 2, in accord with Eq. (2.12). The displacement during a time interval is always equal to the area under the 𝑣𝑥 − 𝑡 curve. This is true even if the acceleration is not constant, although in that case Eq. (2.12) does not apply.  𝟏 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝒕𝟐 (2.12) 𝟐 It’s often useful to have a relationship for position, x-velocity, and (constant) x-acceleration that does not involve time. We have 𝑣𝑥 −𝑣0𝑥 𝑣𝑥 −𝑣0𝑥 𝑎𝑥 = or 𝑡= 𝑡−0 𝑎𝑥 𝑣𝑥 − 𝑣0𝑥 𝟏 𝑣𝑥 − 𝑣0𝑥 𝟐 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 ( ) + 𝒂𝒙 ( ) 𝑎𝑥 𝟐 𝑎𝑥 𝑣𝑥 − 𝑣0𝑥 𝟏 (𝑣𝑥 − 𝑣0𝑥 )𝟐 𝒙 − 𝒙𝟎 = 𝒗𝟎𝒙 + 𝑎𝑥 𝟐 𝒂𝒙 𝟐 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 = 𝟐 𝒗𝟎𝒙 𝑣𝑥 − 𝑣0𝑥 + 𝑣𝑥 − 𝑣0𝑥 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 = 𝟐 𝒗𝟎𝒙 𝑣𝑥 − 𝑣0𝑥 + 𝑣𝑥 − 𝑣0𝑥 𝟐 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 = 𝟐 𝒗𝟎𝒙 𝑣𝑥 − 𝑣0𝑥 + 𝑣𝑥 𝟐 + 𝑣0𝑥 𝟐 −2 𝑣𝑥 𝑣0𝑥 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 = 𝒗𝒙 𝟐 − 𝒗𝟎𝒙 𝟐 𝒗𝒙 𝟐 = 𝒗𝟎𝒙 𝟐 + 𝟐𝒂𝒙 𝒙 − 𝒙𝟎  𝒙−𝒙𝟎 𝒗𝒂𝒗−𝒙 = (2.9) 𝒕 1 𝑣𝑎𝑣−𝑥 = (𝑣0𝑥 +𝑣𝑥 ) (2.10) 2 𝒙 − 𝒙𝟎 1 = (𝑣0𝑥 +𝑣𝑥 ) 𝒕 2 1 𝒙 − 𝒙𝟎 = (𝑣0𝑥 +𝑣𝑥 ) 𝒕 2  𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 𝟏 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝒕𝟐 𝟐 𝒗𝒙 𝟐 = 𝒗𝟎𝒙 𝟐 + 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 1 𝒙 − 𝒙𝟎 = (𝑣0𝑥 +𝑣𝑥 ) 𝒕 2 Freely falling bodies The most familiar example of motion with (nearly) constant acceleration is a body falling under the influence of the earth’s gravitational attraction. Experiment shows that if the effects of the air can be ignored, all bodies at a particular location fall with the same downward acceleration, regardless of their size or weight. The constant acceleration of a freely falling body is called the acceleration due to gravity, and we denote its magnitude with the letter g. We will frequently use the approximate value of g at or near the earth’s surface: The exact value varies with location, so we will often give the value of g at the earth’s surface to only two significant figures as 9.8 m/𝑠 2. On the moon’s surface, the acceleration due to gravity is caused by the attractive force of the moon rather than the earth, and g = 1.6 m/𝑠 2 Near the surface of the sun, g = 270 m/𝑠 2.  𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 𝟏 𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝒕𝟐 𝟐 𝒗𝒙 𝟐 = 𝒗𝟎𝒙 𝟐 + 𝟐𝒂𝒙 𝒙 − 𝒙𝟎 1 𝒙 − 𝒙𝟎 = (𝑣0𝑥 +𝑣𝑥 ) 𝒕 2 Fig. shows our motion diagram for the coin. The motion is vertical, so we use a vertical coordinate axis and call the coordinate y instead of x. We take the origin O at the starting point and the upward direction as positive. 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 𝟏 𝒚 = 𝒚𝟎 + 𝒗𝟎𝒚 𝒕 + 𝒂𝒚 𝒕𝟐 𝟐 Both the initial coordinate 𝒚𝟎 and initial y-velocity 𝒗𝟎𝒚 are zero.