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X. 1D Motion. Sept. 2013. Contents. 1D Motion, Kinematics and Dynamics - definition Speed and Velocity Acceleration Equation of Kinematics Freely falling bodies Graphical analysis of velocity and accelration. 1D Motion, Kinematics and Dynamics.

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## 1D Motion

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**X**1D Motion Sept. 2013**Contents**• 1D Motion, Kinematics and Dynamics - definition • Speed and Velocity • Acceleration • Equation of Kinematics • Freely falling bodies • Graphical analysis of velocity and accelration**1D Motion, Kinematics and Dynamics**One dimensional motion is motion along a straight line, like the motion of a train on a track. There are two aspects to any motion: The movement itself. Is it rapid or slow, for instance?What causes the motion or what changes it? Apply the breaks in a car**Terms (cont.)**Kinematics – deals with the concepts that are needed to describe motion, without any reference to forces Dynamics - deals with the effect that forces have on motion Kinematics and Dynamics form the branch of physics known as Mechanics**Displacement**To describe the motion of an object, we must be able to specify the location of the object at all times Displacement of the car is a vector drawn from the initial position to the final position. Displacement is a vector quantity Final Initial**Term - Displacement**Displacement in one direction along the line is assigned a positive value, and a displacement in the opposite direction is assigned a negative value Displacement does not give information of the location, it gives information about the change in location in specific time.**Average Speed**Usain Bolt run 100 meters in 9.77 seconds last August in Moscow. How fast did he run? 100 m / 9.77 sec = 10.23 m/sec Average speed is the distance traveled divided by the time required to cover the distance (always positive)**Example**My dog chased a cat in our backyard. It run a distance of 30 meters till the cat was lost. The average speed was 6 m/s. How much time did it take till the cat was gone?**Example (cont.)**= 30 m / 6 (m/sec) = 5 sec The dog ran 5 sec till it lost the cat.**Average Velocity**Speed indicates how fast an object is moving. However, speed does not reveal anything about the direction of the motion. To describe both how fast an object moves and the direction of its motion, we need the vector concept of velocity.**Term – Average Velocity**is the initial location at time is the final location at time**Average Velocity – (cont.)**Average velocity is a vector that points in the same direction as the displacement we can represent the direction of motion with a +/– sign Objects A and B have thesame speed s = |v| = +10 m/s, but they have different velocities**We can rewrite the equation as follows:**We can see that the location is defined by the initial location summed up with the multiplication of the average velocity by the time period**V=3 cm/s**X=X0+Vt Example A worm is located 2cm to the right of the origin when measurement start. It crawls towards the origin at constant velocity of 3 cm/sec • Write a function that describe the relation between location and time • Draw a graph of the location at different times • Draw a graph of the average velocity at different times • What is the displacement during 10 sec of movement • What is the distance passed in 10 sec • What is the displacement during the 4th second of movement**X=X0+Vt**• Write a function that describe the relation between location and time • First we decide of the positive direction – it will be to the right • The positive direction impacts the signs of the velocity and the initial location, in this example the velocity will be negative and the location positive V=-3 cm/s X(cm) X0=+2 cm**X=X0+Vt**• Write a function that describe the relation between location and time = 0 X=X0+Vt X=2-3t V=-3 cm/s X(cm) X0=+2 cm**X(cm)**t(s) X=X0+Vt • Draw a graph of the location at different times X=2-3t 2 V=-3 cm/s X(cm) X0=+2 cm**X=X0+Vt**• Draw a graph of the average velocity at different times V(cm/s) t(s) -3 V=-3 cm/s X(cm) X0=+2 cm**X=X0+Vt**• What is the displacement during 10 sec of movement? X=X0+Vt ∆X=Vt X-X0=Vt ∆X=-3∙10=-30cm V=-3 cm/s X(cm) X0=+2 cm**X=X0+Vt**• What is the distance passed in 10 sec? When a body is moving at constant velocity at all times, the distance passed is equal to the absolute value of the displacement S=|∆X|=|-30|=30cm V=-3 cm/s X(cm) X0=+2 cm**X=X0+Vt**• What is the displacement during the 4th second of movement? The 4th second of movement, is 1 second long as any other second. Constant velocity motion has constant displacement at any second -3∙1=-3cm=∆X=Vt V=-3 cm/s X(cm) X0=+2 cm**Instantaneous Velocity**The magnitude of average velocity is an average, hence does not convey any information about how fast you were moving or the direction of the motion at any instant during the trip The instantaneous velocity of the car indicates how fast the car moves and the direction of the motion at each instant of time The magnitude of the instantaneous velocity is called the instantaneous speed**Instantaneous Velocity (cont.)**The notation means that the ratio is defined by a limiting process. Smaller and smaller values of t are used, so small that they approach zero. As smaller t is used, x also becomes smaller. However, the ratio does not become zero,it approaches the value of the instantaneous velocity.**Acceleration**In a wide range of motions, the velocity changes from moment to moment. To describe the manner in which it changes, the concept of acceleration is needed The change in velocity may occur over a short or a long time interval**Acceleration (cont.)**If the velocity is changing, then there is non-zero acceleration**Rewriting the equation:**We received a new function that describes the velocity of a body at any time t, that started off with velocity Vo and travels at constant acceleration a during the period of time t.**V(m/s)**V0 t(s) ΔV Δt Graph of motion in constant acceleration The v vs. t graph slope is the acceleration**V(m/s)**V0 ΔV Δt t(s) Acceleration can be Positive or Negative Negative slope represents negative acceleration**Whenever the acceleration and velocity vectors have opposite**directions, the object slows down and is said to be “decelerating” Is it true to state that any time a body is decelerating, it is also slowing down?**V(m/s)**t(s) ΔV -V0 Δt Negative slope, represent negative acceleration Motion is in negative direction, hence velocity is negative. Attention!!! The magnitude of the velocity is increasing + X**Is it true to state that any time a body have positive**accelerating, it is also speeding up?**Velocity magnitude is decreasing, acceleration is positive**Velocity is negative because motion is towards negative X, and in the same time the velocity is decreasing V(m/s) t(s) Δt ΔV Positive Slope -V0 + X**V(m/s)**ΔV V0 Δt t(s) If velocity is increased while moving in positive direction, both velocity and acceleration are positive + X**V(m/s)**t(s) ΔV V0 Negative Slope => Negative acceleration Δt If velocity is increased while moving in negative direction, both velocity and acceleration are negative Velocity increased to the left a Const. Acceleration left + X**V(m/s)**V0 ΔV Δt If velocity is decreased while moving in positive direction, the velocity is positive, and the acceleration is negative t(s) Velocity decreased to the right Const. Acceleration to left + X**V(m/s)**t(s) Δt ΔV V0 If velocity is decreased while moving in negative direction, the velocity is negative, and the acceleration is positive V Velocity decreased to the left Const. Acceleration to right a + X**Const. negative acceleration**V(m/s) V0 Velocity magnitude decrease Velocity decrease ΔV Δt t(s) Temp stop to switch motion direction Velocity increase Velocity magnitude increase + X**Const. positive acceleration**V(m/s) Velocity magnitude increase ΔV Velocity Increase Δt t(s) Temp stop to switch motion direction Velocity decrease -V0 Velocity magnitude decrease + X**Instantaneous Acceleration**Instantaneous Acceleration is an object’s acceleration at a particular instant of time Instantaneous acceleration is a limiting case of the average acceleration. When the time interval becomes extremely small (approaching zero in the limit), the average acceleration and the instantaneous acceleration become equal**Example 1**The plane in the figure starts from rest (Vo = 0 m/s) when t0 = 0 s. The plane accelerates down the runway and at t=29 s attains a velocity of V=260 km/h, where the plus sign indicates that the velocity points to the right. Determine the average acceleration of the plane?

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