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)). R. Field 1/17/2013 University of Florida PHY 2053 Page 1 1-d Motion: Average Velocity • Average Velocity v v ave The average velocity is defined to be the displacement, Dx, that occurred during a particular interval of time, Dt (i.e. vave = Dx/Dt). Dx Dt x ( t 2 ) x ( t1 ) t 2 t1 x 2 x1 t 2 t1 • 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). R. Field 1/17/2013 University of Florida PHY 2053 Page 2 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) R. Field 1/17/2013 University of Florida shrink Dt t t+Dt t Dt PHY 2053 Page 3 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) R. Field 1/17/2013 University of Florida shrink Dt t+Dt t t Dt PHY 2053 Page 4 1-d Motion: Instantaneous Velocity x x(t+Dt) x(t) R. Field 1/17/2013 University of Florida t+Dt shrink Dt t t Dt PHY 2053 Page 5 1-d Motion: Instantaneous Velocity x x(t+ D t) x(t) R. Field 1/17/2013 University of Florida shrink Dt t t+Dt Dt t PHY 2053 Page 6 1-d Motion: Instantaneous Velocity x tangent line at t x(t) v lim Dt 0 Dx Dt R. Field 1/17/2013 University of Florida dx dt 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 Page 7 1-d Motion: Acceleration • Acceleration a a ave When a particles velocity changes, the particle is said to undergo acceleration (i.e. accelerate). Dv Dt v ( t 2 ) v ( t1 ) t 2 t1 v v 2 v1 t 2 t1 v(t) v2 Hello • Average Acceleration Dv “rise” v1 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 a Dv v The acceleration is the derivative of v(t) with respect to t. a lim Dt 0 R. Field 1/17/2013 University of Florida Dv Dt v(t) dv Instantaneous acceleration a(t) is slope of v-t tangent line at t dt PHY 2053 Page 8 Equations of Motion: a = constant (constant acceleration) • v is a linear function of t a (t ) a Acceleration a (in m/s2) • Special case! Acceleration as a function of time t: a(t) 5 v at t = 0 v v 0 at 4 3 a = 2 m/s2 2 1 0 0 1 2 3 4 5 6 7 8 9 10 time t (in s) Velocity as a function of time t: v(t) x x0 v0t 1 2 at 2 Velocity v (in m/s) • x is a quadratic function of t 20 a = 2 m/s2 v0 = 0 15 10 5 0 0 x at t =0 2 3 4 5 6 7 8 9 10 9 10 time t (in s) Position as a function of time t: x(t) • Note also that 120 a = 2 m/s2 v0 = 0 x0 = 0 2 0 Position x (in m) 100 v v 2 a ( x x0 ) 2 1 80 60 40 20 0 0 1 2 3 4 5 6 7 8 time t (in s) R. Field 1/17/2013 University of Florida PHY 2053 Page 9 Acceleration Due to Gravity • Experimental Result y-axis 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 a y g 9 .8 m / s 2 h The acceleration due to gravity is almost constant and equal to 9.8 m/s2 provided h << RE! x-axis RE v y ( t ) v y 0 gt y (t ) y 0 v y 0 t 1 2 gt Earth 2 v y v y0 2 g ( y y0 ) 2 2 R. Field 1/17/2013 University of Florida PHY 2053 Page 10 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? v t y0 49 m / s g 9 .8 m / s 2 5s Velocity as a function of time t: v(t) 60 a = -9.8 m/s2 v0 = 49 m/s Velocity v (in m/s) 40 What is the ball’s maximum height? 2 h v y0 2g ( 49 m / s ) 2 2 122 . 5 m 20 0 -20 -40 2 (9 .8 m / s ) -60 0 1 2 3 4 5 6 7 8 9 10 time t (in s) How long does it take for the ball to get back to its starting point? 2v y0 10 s 120 g What is the velocity of the ball when it gets back to its starting point? v y v y 0 49 m / s R. Field 1/17/2013 University of Florida PHY 2053 Position y (in m) t Position as a function of time t: y(t) 140 100 80 a = -9.8 m/s2 v0 = 49 m/s y0 = 0 60 40 20 0 0 1 2 3 4 5 6 7 8 9 10 time t (in s) Page 11 2-d Motion: Constant Acceleration • Kinematic Equations of Motion (Vector Form) Acceleration Vector (constant) Velocity Vector (function of t) Position Vector (function of t) The velocity vector and position vector are a function of the time t. a a x xˆ a y yˆ v (t ) v 0 a t r ( t ) r0 v 0 t Warning! These equations are only valid if the acceleration is constant. 1 2 2 at Velocity Vector at time t = 0. v 0 v x 0 xˆ v y 0 yˆ Position Vector at time t = 0. r0 x 0 xˆ y 0 yˆ 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. R. Field 1/17/2013 University of Florida PHY 2053 Page 12 2-d Motion: Constant Acceleration • Kinematic Equations of Motion (Component Form) ax ay constant v x (t ) v x 0 a x t x (t ) x 0 v x 0 t Warning! These equations are only valid if the acceleration is constant. constant v y (t ) v y 0 a y t 1 2 a xt 2 y (t ) y 0 v y 0 t 1 2 a yt 2 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 v x ( t ) v xo 2 a x ( x ( t ) x 0 ) 2 2 v y ( t ) v yo 2 a y ( y ( t ) y 0 ) 2 2 R. Field 1/17/2013 University of Florida PHY 2053 Page 13 Example: Projectile Motion y-axis • Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0. v x ( t ) v 0 cos v y ( t ) v 0 sin gt x ( t ) ( v 0 cos ) t y ( t ) ( v 0 sin ) t v0 x-axis 1 2 gt 2 • Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, t max v 0 sin g H y ( t max ) ( v 0 sin ) 2 2g • Range R (maximum horizontal distance traveled) The time, tf, that it takes the projective reach the ground occurs when y(tf) = 0. Hence, 2 v 0 sin t 2 f 0 y ( t f ) ( v 0 sin ) t f 12 gt f g 2 v 0 sin cos 2 R x ( t f ) ( v 0 cos ) t f R. Field 1/17/2013 University of Florida v 0 sin 2 2 g g PHY 2053 Page 14 Example: Projectile Motion y-axis • Near the Surface of the Earth (h = 0) In this case, ax= 0 and ay= -g, vx0 = v0cos, vy0 = v0sin, x0 = 0, y0 = 0. v x ( t ) v 0 cos v y ( t ) v 0 sin gt x ( t ) ( v 0 cos ) t y ( t ) ( v 0 sin ) t v0 x-axis 1 2 gt 2 • Maximum Height H The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, t max v 0 sin g H y ( t max ) ( v 0 sin ) 2 2g For a fixed v the largest R • Range R (maximum horizontal distance traveled) 0 occurs when = 45o! The time, tf, that it takes the projective reachthe ground occurs when y(tf) = 0. Hence, 2 v 0 sin 2 tf 1 0 y ( t f ) ( v 0 sin ) t f 2 gt f g 2 v 0 sin cos 2 R x ( t f ) ( v 0 cos ) t f R. Field 1/17/2013 University of Florida v 0 sin 2 2 g g PHY 2053 Page 15 Exam 1 Fall 2012: Problem 11 v • Near the surface of the Earth a projectile d is fired from the top of a building at a height h above the ground at an angle h building relative to the horizontal and at a distance d from the edge of the building as shown ground o in the figure. If = 20 and d = 20 m, what 2 is the minimum initial speed, v0, of the v 0 sin 2 d R projectile such that it will make it off the g building and reach the ground? Ignoring air resistance. 0 Answer: 17.5 m/s % Right: 35% v0 R. Field 1/17/2013 University of Florida dg sin 2 2 ( 20 m )( 9 . 8 m / s ) sin( 2 20 ) PHY 2053 17 . 46 m / s 17 . 5 m / s Page 16 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% v x (t ) v 0 x (t ) v 0 t t 2 h v y ( t ) gt y (t ) h 1 2 gt v0 2 2h y-axis g d h d x-axis th Let th be the time the beanbag hits the ground. y (t h ) 0 h 1 2 gt 2 h tan v x (t h ) x (t h ) d v 0 t h R. Field 1/17/2013 University of Florida v y (t h ) gt h v0 76 PHY 2053 2 gt h d 2h 4 d Page 17 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, y-axis h x-axis v y ( t ) gt v x (t ) 0 x (t ) 0 y (t ) h 1 2 gt 2 The time, tf, that the it takes the stone to reach the ground occurs when y(tf) = 0. Hence, 2 1 2h 2 ( 60 m ) y (t f ) 0 h 2 gt f tf 2 g 3 . 494 s ( 9 . 83 m / s ) • 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 In this case, ax= 0 and ay= -g, vx0 = 0, vy0 = -v0, x0 = 0, y0 =h. Hence, y (t ) h v 0 t tf v0 1 2 gt y (t f ) 0 h v 0 t f 2 ( v o ) 2 gh 2 g R. Field 1/17/2013 University of Florida 20 m / s 1 2 gt Quadratic Formula! 2 f ( 20 m / s ) 2 ( 9 . 83 m / s )( 60 m ) 2 9 . 83 m / s PHY 2053 2 2 2 . 01 s Page 18