1/17/2014 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. • Kinematic Equations of Motion (Component Form) r Warning! These a = a x xˆ + a y yˆ equations are only valid if the r r r acceleration is v (t ) = v0 + at constant. r r r r2 1 r (t ) = r0 + v0t + 2 at Velocity Vector at time t = 0. ax = ay = constant v x (t ) = v x 0 + a x t x(t ) = x0 + v x 0t + 12 a xt 2 constant Warning! These equations are only valid if the acceleration is constant. v y (t ) = v y 0 + a y t y (t ) = y0 + v y 0t + 12 a y t 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. r v0 = v x 0 xˆ + v y 0 yˆ r r0 = x0 xˆ + y0 yˆ Position Vector at time t = 0. 2-d Motion: Constant Acceleration The components of the position vector at t = 0, x0 and y0, are constants. • Ancillary Equations Valid at any time t v (t ) − v = 2a x ( x(t ) − x0 ) 2 x The components of the acceleration vector, ax and ay, are constants. 2 xo 2 v y2 (t ) − v yo = 2 a y ( y (t ) − y 0 ) 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. University of Florida PHY 2053 Page 1 University of Florida Projectile Motion PHY 2053 Page 2 Rules of Projectile Motion The x- and y-directions of motion completely independent The x-direction is uniform motion: ax = 0 The y-direction is free fall: ay = -g The initial velocity v 0 can be broken down into its x- and ycomponents v = v cos θ v = v sin θ • Ox x-direction--ax O O Oy O =0 v xo = v o cos θo = v x = constant x = vxot This is the only operative equation in the x-direction since there is O uniform velocity in that direction • y-direction--free fall: a = -g –take the positive direction as upward –uniformly accelerated motion, so the motion equations all hold v = • Velocity at any time Example: Projectile Motion In this case, ax= 0 and ay= -g, vx0 = v0cosθ θ, vy0 = v0sinθ θ, x0 = 0, y0 = 0. v x (t ) = v0 cos θ x(t ) = (v0 cos θ )t v y (t ) = v0 sin θ − gt v0 y (t ) = (v0 sin θ )t − gt θ 2 v x (t ) = v0 cos θ x(t ) = (v0 cos θ )t y-axis y (t ) = (v0 sin θ )t − gt The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, v0 sin θ (v0 sin θ ) 2 tmax = 2g g H = y (tmax ) = 2g For a fixed v the largest R • Range R (maximum horizontal distance traveled) The time, tf, that it takes the projective reach the ground occurs when y(tf) = 0. Hence, 2v sin θ The time, tf, that it takes the projective reachthe ground occurs when y(tf) = 0. Hence, 2v sin θ tf = 0 occurs when θ = 45o! 0 = y (t f ) = (v0 sin θ )t f − 12 gt 2f 0 g University of Florida 0 g 2 0 2 0 PHY 2053 tf = 2v sin θ cos θ v02 sin 2θ R = x(t f ) = (v0 cos θ )t f = = g g 2v sin θ cos θ v sin 2θ R = x(t f ) = (v0 cos θ )t f = = g g 2 0 θ 2 • Range R (maximum horizontal distance traveled) 0 = y (t f ) = (v0 sin θ )t f − 12 gt 2f v0 x-axis 1 2 The time, tmax, that the projective reaches its maximum height occurs when vy(tmax) = 0. Hence, v0 sin θ (v0 sin θ ) 2 H = y (tmax ) = vx v y (t ) = v0 sin θ − gt • Maximum Height H g vy In this case, ax= 0 and ay= -g, vx0 = v0cosθ θ, vy0 = v0sinθ θ, x0 = 0, y0 = 0. • Maximum Height H tmax = θ = tan−1 • Near the Surface of the Earth (h = 0) x-axis 1 2 and Example: Projectile Motion y-axis • Near the Surface of the Earth (h = 0) v x2 + v y2 Page 5 University of Florida PHY 2053 Page 6 1 1/17/2014 Some Special Cases of Projectile Motion Problem-Solving Strategy Select a coordinate system and sketch the path of the projectile--Include initial and final positions, velocities, and accelerations Resolve the initial velocity into x- and ycomponents Treat the horizontal and vertical motions independently • Horizontal motion: Use techniques for problems with constant velocity Object may be fired horizontally The initial velocity is all in the x-direction • vo = vx and vy = 0 All the general rules of projectile motion apply • Vertical motion: Use techniques for problems with constant acceleration Exam 1 Fall 2012: Problem 11 Exam 1 Spring 2012: Problem 12 v • Near the surface of the Earth a projectile θ d is fired from the top of a building at a height h above the ground at an angle θ h building relative to the horizontal and at a distance d from the edge of the building as shown ground in the figure. If θ = 20o and d = 20 m, what is the minimum initial speed, v0, of the v02 sin 2θ projectile such that it will make it off the d = R = g building and reach the ground? Ignoring air resistance. 0 Answer: 17.5 m/s % Right: 35% Let th be the time the beanbag hits the ground. dg ( 20m)(9.8m / s 2 ) v0 = = ≈ 17.46m / s ≈ 17.5m / s sin 2θ sin(2 × 20o ) University of Florida PHY 2053 Relative Position Vector Position of car A relative to car B is given by the vector subtraction equation r • rAE is the position of car A as measured by E r • rBE is the position of car B as measured by E r • rABis the position of car A as measured by car B r r r rAB = rAE − rEB • 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. 2h y-axis Answer: 76.0°below the horizontal t h2 = % Right: 22% g h v y (t ) = − gt v x (t ) = v0 d d x-axis v0 = 2 1 θ y ( t ) = h − gt th x(t ) = v0t 2 Page 9 y (t h ) = 0 = h − 12 gth2 x(t h ) = d = v0th University of Florida tan θ = v y (t h ) v x (t h ) gt h gt h2 2h = = =4 v0 d d o θ ≈ 76 = PHY 2053 Page 10 Relative Velocity Notation The pattern of subscripts can be useful in solving relative velocity problems Assume the following notation: • E is an observer, stationary with respect to the earth • A and B are two moving cars The rate of change of the displacements gives the relationship for the velocities r r r v AB = v AE − v EB 2