R. Field 1/17/2013 University of Florida PHY 2053Page 1 1-d Motion: Position & Displacement 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. The x-axis: Displacement: The change from position x 1 to position x 2 is called the displacement,  x.  x = x 2 –x 1 Graphical Technique: The displacement has both a magnitude, |  x|, and a direction (positive or negative). 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)). time t x(t)

R. Field 1/17/2013 University of Florida PHY 2053Page 2 1-d Motion: Average Velocity Average Velocity The average velocity is defined to be the displacement,  x, that occurred during a particular interval of time,  t (i.e. v ave =  x/  t). Average Speed The average speed is defined to be the magnitude of total distance covered during a particular interval of time,  t (i.e. s ave = (total distance)/  t).

R. Field 1/17/2013 University of Florida PHY 2053Page 3 1-d Motion: Instantaneous Velocity t t+  t  tt x(t+  t) x(t) shrink  t t x

R. Field 1/17/2013 University of Florida PHY 2053Page 4 1-d Motion: Instantaneous Velocity t t+  t  tt x(t+  t) x(t) shrink  t t x

R. Field 1/17/2013 University of Florida PHY 2053Page 5 1-d Motion: Instantaneous Velocity t t+  t  tt x(t+  t) x(t) shrink  t t x

R. Field 1/17/2013 University of Florida PHY 2053Page 6 1-d Motion: Instantaneous Velocity t t+  t  tt x(t+  t) x(t) shrink  t t x

R. Field 1/17/2013 University of Florida PHY 2053Page 7 1-d Motion: Instantaneous Velocity t x(t) Instantaneous velocity v(t) is slope of x-t tangent line at t tangent line at t t x The velocity is the derivative of x(t) with respect to t.

R. Field 1/17/2013 University of Florida PHY 2053Page 8 1-d Motion: Acceleration Average Acceleration The average acceleration is defined to be the change in velocity,  v, that occurred during a particular interval of time,  t (i.e. a ave =  v/  t). Instantaneous Acceleration The acceleration is the derivative of v(t) with respect to t. Acceleration When a particles velocity changes, the particle is said to undergo acceleration (i.e. accelerate). v1v1 v2v2 v v(t)  v “rise” a vv v v(t) Instantaneous acceleration a(t) is slope of v-t tangent line at t

R. Field 1/17/2013 University of Florida PHY 2053Page 9 Equations of Motion: a = constant Special case! (constant acceleration) v is a linear function of t x is a quadratic function of t v at t = 0 x at t =0 Note also that

R. Field 1/17/2013 University of Florida PHY 2053Page 10 Acceleration Due to Gravity Experimental Result Earth y-axis x-axis RERE h Near the surface of the Earth all objects fall toward the center of the Earth with the same constant acceleration, g ≈ 9.8 m/s 2, (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/s 2 provided h << R E !

R. Field 1/17/2013 University of Florida PHY 2053Page 11 Equations of Motion: Example Problem 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. y-axis v y0 = 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?

R. Field 1/17/2013 University of Florida PHY 2053Page 12 2-d Motion: Constant Acceleration Kinematic Equations of Motion (Vector Form) The velocity vector and position vector are a function of the time t. Acceleration Vector (constant) Velocity Vector (function of t) Position Vector (function of t) Velocity Vector at time t = 0. Position Vector at time t = 0. The components of the acceleration vector, a x and a y, are constants. The components of the velocity vector at t = 0, v x0 and v y0, are constants. The components of the position vector at t = 0, x 0 and y 0, are constants. Warning! These equations are only valid if the acceleration is constant.

R. Field 1/17/2013 University of Florida PHY 2053Page 13 2-d Motion: Constant Acceleration Kinematic Equations of Motion (Component Form) constant The components of the acceleration vector, a x and a y, are constants. The components of the velocity vector at t = 0, v x0 and v y0, are constants. The components of the position vector at t = 0, x 0 and y 0, are constants. Warning! These equations are only valid if the acceleration is constant. constant Ancillary Equations Valid at any time t

R. Field 1/17/2013 University of Florida PHY 2053Page 14 Example: Projectile Motion Near the Surface of the Earth (h = 0) In this case, a x = 0 and a y = -g, v x0 = v 0 cos , v y0 = v 0 sin , x 0 = 0, y 0 = 0. Maximum Height H The time, t max, that the projective reaches its maximum height occurs when v y (t max ) = 0. Hence, Range R (maximum horizontal distance traveled) The time, t f, that it takes the projective reach the ground occurs when y(t f ) = 0. Hence,

R. Field 1/17/2013 University of Florida PHY 2053Page 15 Example: Projectile Motion Near the Surface of the Earth (h = 0) In this case, a x = 0 and a y = -g, v x0 = v 0 cos , v y0 = v 0 sin , x 0 = 0, y 0 = 0. Maximum Height H The time, t max, that the projective reaches its maximum height occurs when v y (t max ) = 0. Hence, Range R (maximum horizontal distance traveled) The time, t f, that it takes the projective reachthe ground occurs when y(t f ) = 0. Hence, For a fixed v 0 the largest R occurs when  = 45 o !

R. Field 1/17/2013 University of Florida PHY 2053Page 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  relative to the horizontal and at a distance d from the edge of the building as shown in the figure. If  = 20 o and d = 20 m, what is the minimum initial speed, v 0, 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%

R. Field 1/17/2013 University of Florida PHY 2053Page 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 t h be the time the beanbag hits the ground.

R. Field 1/17/2013 University of Florida PHY 2053Page 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/s 2. If you drop the stone how long does it take for it to fall to the base of the gorge? In this case, a x = 0 and a y = -g, v x0 = 0, v y0 = 0, x 0 = 0, y 0 =h. Hence, The time, t f, that the it takes the stone to reach the ground occurs when y(t f ) = 0. Hence, If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground? In this case, a x = 0 and a y = -g, v x0 = 0, v y0 = -v 0, x 0 = 0, y 0 =h. Hence, Have to use the Quadratic Formula!