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Kinematics: Motion in One Dimension 2.1 Displacement & Velocity Learning Objectives • Describe motion in terms of displacement, time, and velocity • Calculate the displacement of an object traveling at a known velocity for a specific time interval • Construct and interpret graphs of position versus time Essential Concepts • • • • Frames of reference Vector vs. scalar quantities Displacement Velocity – Average velocity – Instantaneous velocity • Acceleration • Graphical representation of motion Reference Frames • Motion is relative • When we say an object is moving, we mean it is moving relative to something else (reference frame) Scalar Quantities & Vector Quantities • Scalar quantities have magnitude • Example: speed 15 m/s • Vector quantities have magnitude and direction • Example: velocity 15 m/s North Displacement • Displacement is a vector quantity • Indicates change in location (position) of a body ∆x = xf - xi • It is specified by a magnitude and a direction. • Is independent of the path traveled by an object. Displacement is change in position www.cnx.org Displacement vs. Distance • Distance is the length of the path that an object travels • Displacement is the change in position of an object Describing Motion Describing motion requires a frame of reference http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html Determining Displacement In these examples, position is determined with respect to the origin, displacement wrt x1 http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html Indicating Direction of Displacement Direction can be indicated by sign, degrees, or geographical directions. When sign is used, it follows the conventions of a standard graph Positive Right Up Negative Left Down Displacement • Linear change in position of an object • Is not the same as distance Displacement • • • • Distance = length (blue) How many units did the object move? Displacement = change in position (red) How could you calculate the magnitude of line AB? • ≈ 5.1 units, NE Reference Frames & Displacement • Direction is relative to the initial position, x1 • x1 is the reference point Average Velocity Speed: how far an object travels in a given time interval Velocity includes directional information: Average Velocity displacement average velocity = time interval Dx x f - xi v= = Dt t Velocity • Example • A squirrel runs in a straight line, westerly direction from one tree to another, covering 55 meters in 32 seconds. Calculate the squirrel’s average velocity • vavg = ∆x / ∆t • vavg = 55 m / 32 s • vavg = 1.7 m/s west Velocity can be represented graphically: Position Time Graphs Velocity can be interpreted graphically: Position Time Graphs Find the average velocity between t = 3 min to t = 8 min Calculate the average velocity for the entire trip Formative Assessment: Position-Time Graphs Object at rest? Traveling slowly in a positive direction? Traveling in a negative direction? Traveling quickly in a positive direction? dev.physicslab.org Average vs. Instantaneous Velocity • Velocity at any given moment in time or at a specific point in the object’s path Position-time when velocity is not constant Average velocity compared to instantaneous velocity Instantaneous velocity is the slope of the tangent line at any particular point in time. Instantaneous Velocity • The instantaneous velocity is the average velocity, in the limit as the time interval becomes infinitesimally short. 2.2 Acceleration 2.2 Acceleration Learning Objectives • Describe motion in terms of changing velocity • Compare graphical representations of accelerated and non-accelerated motions • Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration X-t graph when velocity is changing Acceleration Acceleration is the rate of change of velocity. Acceleration: Change in Velocity • Acceleration is the rate of change of velocity • a = ∆v/∆t • a = (vf – vi) / (tf – ti) • Since velocity is a vector quantity, velocity can change in magnitude or direction • Acceleration occurs whenever there is a change in magnitude or direction of movement. Acceleration Because acceleration is a vector, it must have direction Here is an example of negative acceleration: Customary Dimensions of Acceleration • a = ∆v/∆t • = m/s/s • = m/s2 • Sample problems 2B A bus traveling at 9.0 m/s slows down with an average acceleration of -1.8 m/s. How long does it take to come to a stop? Negative Acceleration • Both velocity & acceleration can have (+) and (-) values • Negative acceleration does not always mean an object is slowing down Is an object speeding up or slowing down? • Depends upon the signs of both velocity and acceleration Velocity + + - Accel + + Motion Speeding up in + dir Speeding up in - dir Slowing down in + dir Slowing down in - dir • Construct statement summarizing this table. Velocity-Time Graphs • Is this object accelerating? • How do you know? • What can you say about its motion? www.gcsescience.com Velocity-Time Graph • • • • Is this object accelerating? How do you know? What can you say about its motion? What feature of the graph represents acceleration? www.gcsescience.com Velocity-Time Graph dev.physicslab.org Displacement with Constant Acceleration (C) x vavg t Since Then Thus Or and vavg vi v f vavg vavg x vi v f t 2 vi v f x 2 t 1 x vi v f t 2 2 Displacement on v-t Graphs How can you find displacement on the v-t graph? x v , so x vt t Displacement on v-t Graphs x vt Displacement is the area under the line! Graphical Representation of Displacement during Constant Acceleration Displacement on a Non-linear v-t graph • If displacement is the area under the v-t graph, how would you determine this area? Final velocity of an accelerating object v Since a t v f vi and a t Then v f vi at Displacement During Constant Acceleration (D) 1 We know : Δx vi v f Δt 2 We know : v f vi aΔt Substituti ng v f into the above equation : 1 Δx vi vi aΔt Δt 2 1 Δx 2vi aΔt Δt 2 1 Δx vi Δt aΔt 2 2 Graphical Representation Derivation of the Equation Final velocity after any displacement (E) v f vi 2ax 2 2 A baby sitter pushes a stroller from rest, accelerating at 0.500 m/s2. Find the velocity after the stroller travels 4.75m. (p. 57) Identify the variables. Solve for the unknown. Substitute and solve. Kinematic Equations x x f xi vavg 1 x (vi v f )t 2 1 2 x vi t at 2 x t v a t v f vi at v f vi 2ax 2 2 2.3 Falling Objects Objectives 1. Relate the motion of a freely falling body to motion with constant acceleration. 2. Calculate displacement, velocity, and time at various points in the motion of a freely falling object. 3. Compare the motions of different objects in free fall. Motion Graphs of Free Fall What do motion graphs of an object in free fall look like? Motion Graphs of Free Fall What do motion graphs of an object in free fall look like? x-t graph v-t graph Do you think a heavier object falls faster than a lighter one? Why or why not? Yes because …. No, because …. Free Fall • In the absence of air resistance, all objects fall to earth with a constant acceleration • The rate of fall is independent of mass • In a vacuum, heavy objects and light objects fall at the same rate. • The acceleration of a free-falling object is the acceleration of gravity, g • g = 9.81m/s2 memorize this value! Free Fall • Free fall is the motion of a body when only the force due to gravity is acting on the body. • The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration. • Free-fall acceleration is denoted with by ag (generally) or g (on Earth’s surface). Free Fall Acceleration • Free-fall acceleration is the same for all objects, regardless of mass. • This book will use the value g = 9.81 m/s2. • Free-fall acceleration on Earth’s surface is – 9.81 m/s2 at all points in the object’s motion. • Consider a ball thrown up into the air. – Moving upward: velocity is decreasing, acceleration is –9.81 m/s2 – Top of path: velocity is zero, acceleration is –9.81 m/s2 – Moving downward: velocity is increasing, acceleration is –9.81 m/s2 Sample Problem • Falling Object • A player hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. • If the volleyball starts from 2.0 m above the floor, • how long will it be in the air before it strikes the floor? Sample Problem, continued 1. Define Given: vi = +6.0 m/s a = –g = –9.81 m/s2 Δ y = –2.0 m Diagram: Place the origin at the Starting point of the ball (yi = 0 at ti = 0). Unknown: Δt = ? 2. Plan Choose an equation or situation: Both ∆t and vf are unknown. v f vi 2ay 2 2 v f vi at We can determine ∆t if we know vf Solve for vf then substitute & solve for ∆t 3. Calculate Rearrange the equation to isolate the unknowns: v f vi 2ay 2 vf = - 8.7 m/s t v f vi a Δt = 1.50 s Summary of Graphical Analysis of Linear Motion This is a graph of x vs. t for an object moving with constant velocity. The velocity is the slope of the x-t curve. Comparison of v-t and x-t Curves On the left we have a graph of velocity vs. time for an object with varying velocity; on the right we have the resulting x vs. t curve. The instantaneous velocity is tangent to the curve at each point. Displacement an v-t Curves The displacement, x, is the area beneath the v vs. t curve.