Chapter 4 Motion in Two Dimensions Position Vector r r f − ri Velocity Vector v avg r t r d r v lim = t → 0 t dt v= v Acceleration Vector a avg v dv = t → 0 t dt a lim v v f − v i = t t f − ti Two-Dimensional Motion with Constant Acceleration r = xˆi + yˆj Two-Dimensional Motion with Constant Acceleration v= dr dx ˆ dy ˆ = i+ j = vx ˆi + v y ˆj dt dt dt vxf = vxi + ax t and v yf = v yi + a y t ( ) ( ) v f = ( vxi + ax t ) ˆi + ( v yi + a y t ) ˆj = vxi ˆi + v yi ˆj + ax ˆi + a y ˆj t v f = v i + at ( for constant a ) Two-Dimensional Motion with Constant Acceleration 1 1 x f = xi + vxi t + ax t 2 y f = yi + v yi t + a y t 2 2 2 1 1 r f = xi + vxi t + ax t 2 ˆi + yi + v yi t + a y t 2 ˆj 2 2 1 = xi ˆi + yi ˆj + vxi ˆi + v yi ˆj t + ax ˆi + a y ˆj t 2 2 ( ) ( 1 r f = ri + v i t + at 2 2 ) ( ( for constant a ) ) Two-Dimensional Motion with Constant Acceleration v f = v i + at 1 r f = ri + v i t + at 2 2 ( for constant a ) Example 4.1: Motion in a Plane A particle moves in the xy plane, starting from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of −15 m/s. The particle experiences an acceleration in the x direction, given by ax = 4.0 m/s2. (A) Determine the total velocity vector at any later time. Example 4.1: Motion in a Plane Example 4.1: Motion in a Plane (B) Calculate the velocity and speed of the particle at t = 5.0 s and the angle the velocity vector makes with the x axis. Example 4.1: Motion in a Plane (C) Determine the x and y coordinates of the particle at any time t and its position vector at this time. Projectile Motion Projectile Motion 1 r f = ri + v i t + gt 2 2 vxi = vi cos i v yi = vi sin i Projectile Motion x f = xi + vxi t v yf = v yi − gt v y ,avg = v yi + v yf 2 1 y f = yi + ( v yi + v yf ) t 2 1 y f = yi + v yi t − gt 2 2 v yf 2 = v yi 2 − 2 g ( y f − yi ) Horizontal Range and Maximum Height of a Projectile Height of a Projectile v yf = v yi − gt → 0 = vi sin i − gt A → t A = vi sin i y f = yi + v yi t − g 1 2 gt → 2 v sin i 1 vi sin i h = ( vi sin i ) i − g g 2 g vi 2 sin 2 i h= 2g 2 Horizontal Range of a Projectile x f = xi + vxi t → R = vxi t B = ( vi cos ) 2t A 2vi sin i 2vi 2 sin i cos i = ( vi cos i ) = g g 2sin i cos i = sin 2i vi 2 sin 2i R= g Rmax vi 2 = g Trajectories of a Projectile Example 4.2: The Long Jump A long jumper leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0 m/s. (A) How far does he jump in the horizontal direction? Example 4.2: The Long Jump (B) What is the maximum height reached? Example 4.3: A Bull’s-Eye Every Time In a popular lecture demonstration, a projectile is aimed directly at a target in such a way that the projectile leaves the gun at the same time the target is dropped from rest. Show that the projectile hits the falling target as shown in the figure. Example 4.3: A Bull’s-Eye Every Time Example 4.4: That’s Quite an Arm A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of 20.0 m/s as shown in the figure. The height from which the stone is thrown is 45.0 m above the ground. (A) How long does it take the stone to reach the ground? Example 4.4: That’s Quite an Arm Example 4.4: That’s Quite an Arm (B) What is the speed of the stone just before it strikes the ground? Example 4.5: The End of the Ski Jump A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in the figure. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline? Example 4.5: The End of the Ski Jump Analysis Model: Particle in Uniform Circular Motion dv a= dt Analysis Model: Particle in Uniform Circular Motion v v = r r a avg = v t = v r r t Centripetal Acceleration v2 ac = r Particle in Uniform Circular Motion v= 2 r 2 r → T= v T 2 = T v v = 2 = → v = r 2 r r ac = ( r ) r 2 = r 2 Analysis Model: Particle in Uniform Circular Motion v2 ac = r 2 r T= v 2 = T Example 4.6: The Centripetal Acceleration of the Earth (A) What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? Example 4.6: The Centripetal Acceleration of the Earth (B) What is the angular speed of the Earth in its orbit around the Sun? Tangential and Radial Acceleration a = a r + at dv at = dt v2 ar = −ac = − r a = ar 2 + at 2 Example 4.7: Over the Rise A car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant? Example 4.7: Over the Rise