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1-d Motion: Position & Displacement

1-d Motion: Position & Displacement. The x-axis:. We locate objects by specifying their position along an axis (in this case x-axis). The positive direction of an axis is in the direction of increasing numbers. The opposite is the negative direction. Displacement:. x(t).

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1-d Motion: Position & Displacement

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  1. 1-d Motion: Position & Displacement • The x-axis: We locate objects by specifying their position along an axis (in this case x-axis). The positive direction of an axis is in the direction of increasing numbers. The opposite is the negative direction. • Displacement: x(t) The change from position x1 to position x2 is called the displacement, Dx. Dx = x2 –x1 The displacement has both a magnitude, |Dx|, and a direction (positive or negative). time t • Graphical Technique: A convenient way to describe the motion of a particle is to plot the position x as a function of time t (i.e. x(t)). PHY 2053

  2. 1-d Motion: Average Velocity • Average Velocity The average velocity is defined to be the displacement, Dx, that occurred during a particular interval of time, Dt (i.e. vave = Dx/Dt). • Average Speed The average speed is defined to be the magnitude of total distance covered during a particular interval of time, Dt (i.e. save = (total distance)/Dt). PHY 2053

  3. 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t)  t t+Dt t shrinkDt Dt PHY 2053

  4. 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) t  t+Dt t shrinkDt Dt PHY 2053

  5. 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) t  t t+Dt shrinkDt Dt PHY 2053

  6. 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) t  t t+Dt shrinkDt Dt PHY 2053

  7. 1-d Motion: Instantaneous Velocity x tangent line at t x(t) Instantaneous velocity v(t) is slope of x-t tangent line at t t t The velocity is the derivative of x(t) with respect to t. PHY 2053

  8. 1-d Motion: Acceleration v v(t) v2 Dv “rise” v1 Dv a • Acceleration When a particles velocity changes, the particle is said to undergo acceleration (i.e. accelerate). • Average Acceleration The average acceleration is defined to be the change in velocity, Dv, that occurred during a particular interval of time, Dt (i.e. aave = Dv/Dt). • Instantaneous Acceleration v The acceleration is the derivative of v(t) with respect to t. v(t) Instantaneous acceleration a(t) is slope of v-t tangent line at t PHY 2053

  9. Equations of Motion: a = constant • Special case! (constant acceleration) v at t = 0 • v is a linear function of t • x is a quadratic function of t x at t =0 • Note also that PHY 2053

  10. Acceleration Due to Gravity y-axis h x-axis RE Earth • Experimental Result Near the surface of the Earth all objects fall toward the center of the Earth with the same constant acceleration, g ≈ 9.8 m/s2, (in a vacuum) independent of mass, size, shape, etc. • Equations of Motion The acceleration due to gravity is almost constant and equal to 9.8 m/s2 provided h << RE! PHY 2053

  11. Equations of Motion: Example Problem y-axis • Example Problem A ball is tossed up along the y-axis (in a vacuum on the Earth’s surface) with an initial speed of 49 m/s. vy0 = 49 m/s Earth How long does the ball take to reach its maximum height? What is the ball’s maximum height? How long does it take for the ball to get back to its starting point? What is the velocity of the ball when it gets back to its starting point? PHY 2053

  12. 2-d Motion: Constant Acceleration • Kinematic Equations of Motion (Vector Form) Acceleration Vector (constant) Warning! These equations are only valid if the acceleration is constant. Velocity Vector (function of t) Position Vector (function of t) The velocity vector and position vector are a function of the time t. Velocity Vector at time t = 0. Position Vector at time t = 0. The components of the acceleration vector, ax and ay, are constants. The components of the velocity vector at t = 0, vx0 and vy0, are constants. The components of the position vector at t = 0, x0 and y0, are constants. PHY 2053

  13. 2-d Motion: Constant Acceleration • Kinematic Equations of Motion (Component Form) Warning! These equations are only valid if the acceleration is constant. constant constant The components of the acceleration vector, ax and ay, are constants. The components of the velocity vector at t = 0, vx0 and vy0, are constants. The components of the position vector at t = 0, x0 and y0, are constants. • Ancillary Equations Valid at any time t PHY 2053

  14. Example: Projectile Motion • Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cosq, vy0 = v0sinq, x0 = 0, y0 = 0. • Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, • Range R (maximum horizontal distance traveled) The time, tf, that it takes the projective reach the ground occurs when y(tf) = 0. Hence, PHY 2053

  15. Example: Projectile Motion • Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cosq, vy0 = v0sinq, x0 = 0, y0 = 0. • Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, • Range R (maximum horizontal distance traveled) For a fixed v0 the largest R occurs when q = 45o! The time, tf, that it takes the projective reachthe ground occurs when y(tf) = 0. Hence, PHY 2053

  16. Exam 1 Fall 2012: Problem 11 • Near the surface of the Earth a projectile is fired from the top of a building at a height h above the ground at an angle q relative to the horizontal and at a distance d from the edge of the building as shown in the figure. If q = 20o and d = 20 m, what is the minimum initial speed, v0, of the projectile such that it will make it off the building and reach the ground? Ignoring air resistance. Answer: 17.5 m/s % Right: 35% PHY 2053

  17. Exam 1 Spring 2012: Problem 12 • A beanbag is thrown horizontally from a dorm room window a height h above the ground. It hits the ground a horizontal distance d = h/2 from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag's velocity just before impact. Answer: 76.0° below the horizontal % Right: 22% Let th be the time the beanbag hits the ground. PHY 2053

  18. Example Problem: Projectile Motion • A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped off the bridge. Ignore air resistance. At the location of the bridge g has been measured to be 9.83 m/s2. If you drop the stone how long does it take for it to fall to the base of the gorge? In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = 0, x0 = 0, y0 =h. Hence, The time, tf, that the it takes the stone to reach the ground occurs when y(tf) = 0. Hence, • If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground? Have to use the Quadratic Formula! In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = -v0, x0 = 0, y0 =h. Hence, PHY 2053

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